Current Electricity
Solutions for Physics, Class 10, ICSE
Exercise 8A Long Questions
7 questionsAnswer:
According to Ohm's law, the current flowing in a conductor is directly proportional to the potential difference applied across it's ends provided that the physical conditions and the temperature of the conductor remain constant.
If a current I flows in a conductor when the potential difference across it's ends is V and R is the resistance, then according to Ohm's law V = IR
Below circuit diagram shows the setup for verifying Ohm's law:

Answer:
The conductors which obey the Ohm's law are called the ohmic resistors or linear resistances. Examples are all metallic conductors (such as Silver, Aluminium, Copper, Iron etc.)
Current–Voltage relationship of an ohmic resistor is shown in the below graph:

Resistance is determined in the form of slope from the graph.
Answer:
The conductors which do not obey the Ohm's law are known as the non-ohmic resistors (or non-linear resistances). Examples — LED, solar cell, junction diode, etc.
Current-Voltage relationship of non-ohmic resistors is shown in the graph below:

Answer:
V-I graph of a conductor at two different temperatures T1 and T2 is shown below:

Figure shows two straight lines A and B on the V-I graph for a conductor at two different temperatures T1 and T2 (T1 > T2) respectively.
The straight line A is more steeper than the line B because the resistance of conductor is more at high temperature T1 than at low temperature T2
Answer:
(a) In a thick conductor, electrons get a larger area of cross section to flow as compared to a thin conductor, therefore a thick wire offers less resistance. The resistance of a conductor is inversely proportional to it's area of cross section (a), normal to the direction of flow of current i.e.,
R ∝
Hence,
R ∝
where r is the radius of the wire.
Hence, resistance of a wire is inversely proportional to the square of the radius of the wire.
(b) As we have seen above that the resistance of a conductor is inversely proportional to it's area of cross section (a), normal to the direction of flow of current. Hence, the thin wire, will offer more resistance as it's area of cross section is less.
Exercise 8A Multiple Choice Type
20 questionsAnswer:
ammeter, series
Reason — By connecting the ammeter in series, the entire current flowing through the circuit also flows through the ammeter, allowing it to measure the current accurately.
Answer:
V = W/Q
Reason — If W joule of work is done in bringing a test charge Q coulomb from infinity to the point P, then electric potential V at point P is V = W/Q .
Answer:
1 joule
Reason — The potential at a point is defined as 1 volt when 1 joule of work is done in bringing 1 coulomb charge from infinity to that point.
Answer:
voltmeter, parallel
Reason — Connecting the voltmeter in parallel ensures that it measures the potential difference accurately because it doesn't affect the current flow in the circuit.
Answer:
solar cell
Reason — Nichrome, Copper sulphate solution with copper electrodes and dil sulphuric acid, Silver, all the three follow Ohm's Law i.e., V = IR, hence they are ohmic conductors whereas a solar cell is a non-ohmic conductor.
Answer:
all of the above
Reason — The three factors on which the resistance of wire depends are —
- Dependence on length of the wire — the resistance of a wire is directly proportional to the length of the wire.
R ∝ l - Dependence on the thickness of the wire — the resistance of a wire is inversely proportional to it's area of cross section (a) normal to the direction of flow of current.
R ∝ - Dependence on the temperature of the wire — the resistance of conductor increases with an increase in it's temperature.
Answer:
Same
Reason — Specific resistance of an object is an intrinsic property which is independent of it's dimensions but does depend upon nature of material and temperature. Since both rod and thin wire are of same material i.e., aluminium so both will have same specific resistance.
Answer:
9R
Reason — When the wire is stretched to thrice it's length, it's area of cross section becomes 1/3 and it's length becomes 3l.
Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.
Then,
length = l,
Resistance = R Ω,
new length = 3l,
new area =
From relation
R = ρ = ρ
Initial resistance R1 = R = ρ [Equation 1]
New resistance Rn = ρ = ρ [Equation 2]
On dividing eqn (ii) by (i), we get,
Hence, the new resistance = 9R.
Answer:
nichrome wire
Reason — The conductors which obey the Ohm's law are called the ohmic resistors or linear resistances. A Nichrome wire obeys Ohm's law hence it is an ohmic resistor.
Answer:
Both (1) and (3)
Reason — Nichrome wire is commonly used in heating appliances due to its following properties:
- Specific resistance is high — Nichrome wire has a relatively high specific resistance, which means it resists the flow of electric current efficiently, resulting in heating when current passes through it.
- Resistance increases with an increase in temperature — This property is desirable for heating elements because it helps maintain a consistent level of heating even as the temperature rises.
Answer:
both (1) and (3)
Reason — A superconductor exhibits both infinite conductance and zero resistance, allowing electric current to flow through it without any loss of energy due to resistance.
Exercise 8A Numericals
13 questionsAnswer:
Given,
Number of electrons flowing (n) = 6.25 × 1016
Time taken (t) = 2 s
e = 1.6 × 10-19 C
Current in conductor is given by,
I =
Substituting the values in the formula above we get,
Hence, the current flowing through the conductor = 5 mA from B to A
Answer:
Given,
Current (I) = 3.2 mA = 3.2 × 10-3 A
Charge of one electron = -1.6 × 10-19 coulomb
t = 1 s
Charge flowing through the conductor in one second
Charge (Q) = current (I) x time (t)
Substituting the values in the formula above we get,
Q = 3.2 x 10-3 x 1 = 3.2 x 10-3
Hence,
Therefore, the number of electrons that will pass each second through the cross section of that conductor = 2 x 1016
Answer:
Given,
Current (I) = 300 mA = 0.3 A
Resistance (R) = 20 Ω
Potential difference (V) = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
V = 0.3 × 20
V = 6 V
Hence, the required potential difference = 6 V
Answer:
Given,
Current (I) = 1.2 A
Potential difference (V) = 6.0 V
Resistance (R) = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
6 = 1.2 x R
⇒ R =
⇒ R = 5 Ω
Hence, resistance of filament of bulb = 5 Ω
Answer:
Given,
Potential difference (V) = 12 V
Current (I) = 2 A
Resistance = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
12 = 2 x R
⇒ R =
⇒ R = 6 Ω
Hence, the resistance of the filament of the bulb = 6 Ω
The resistance will be less when the bulb is not glowing.
Answer:
Given,
Resistance (R) = 5 Ohm
Potential difference (V) = 3 V
Current (I) = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
3 = I x 5
⇒ I =
⇒ I = 0.6 A
Hence, the current flowing through the wire = 0.6 A
In an experiment of verification of Ohm's law, following observations are obtained.
Potential difference V (in volt) | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 |
---|---|---|---|---|---|
Current I (in amp) | 0.2 | 0.4 | 0.6 | 0.8 | 1.0 |
Draw a V-I graph and use this graph to find —
(a) the potential difference V when the current I is 0.5 A.
(b) the current I when the potential difference V is 0.75 V.
(c) the resistance in circuit.
Answer:
V-I graph for the readings is shown below:

(a) We can observe from the graph that when current is 0.5 A then potential difference is 1.25 V.
(b) We can observe from the graph that when potential difference is 0.75 V, then current is 0.3 A.
(c) Slope of the graph gives the resistance, and as the graph is linear
So, if potential difference (V) = 1.0 V
current (I) = 0.4 A
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
1.0 = 0.4 x R
⇒ R =
⇒ R =
⇒ R = 2.5 Ω
Hence, the resistance in circuit = 2.5 Ω
Answer:
(i) R = ρ = ρ
Resistance for wire of radius 1 mm,
R1 = ρ = ρ
Resistance for wire of radius 2 mm,
R2 = ρ = ρ
Hence, ratio between the two,
R1 : R2
Hence, the ratio of resistance between the two wires = 4 : 1
(ii) The specific resistance of the two wires will be same because the specific resistance is a characteristic property of the material, hence it does not change with radius.
Therefore, ratio of specific resistance between the two wires = 1 : 1
Answer:
When the wire is stretched to double it's length, it's area of cross section becomes half and it's length becomes double.
Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.
Then,
length = l,
R = 1 Ω,
new length = 2l,
new area =
From relation
R = ρ = ρ
Initial resistance R1 = 1 = ρ [Equation 1]
New resistance Rn = ρ = ρ [Equation 2]
On dividing eqn (ii) by (i), we get,
Hence, the new resistance = 4 Ω.
Answer:
Given,
Resistance (R) = 3 Ω
Length (l) = 10 cm
Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.
Then,
R = 3 Ω
length = 10 cm,
new length = 30 cm,
new area a =
From relation
R = ρ = ρ
Initial resistance 3 = ρ [Equation 1]
New resistance R2 = ρ = ρ [Equation 2]
On dividing eqn (i) by (ii), we get,
Hence, the new resistance = 27 Ω.
Answer:
Given,
Resistance (R) = 9 Ω
When the wire is tripled on itself, it's area of cross section becomes thrice and it's length becomes .
Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.
Then,
length = 30 cm,
new length = = = 10 cm,
new area an = 3a
From relation
R = ρ
Initial resistance 9 = ρ [Equation 1]
New resistance Rn = ρ [Equation 2]
On dividing eqn (ii) by (i), we get,
Hence, the new resistance = 1 Ω.
Answer:
Given,
Specific resistance ρ = 1.7 × 10-8Ω m
Radius r = 1 mm = 10-3m
Resistance R = 2 Ω
Length l = ?
From relation
R = ρ = ρ
Substituting the values in the formula above we get,
Hence, the length of the copper wire = 369.4 m.
Answer:
Case 1 —
Current (I) = 100 m A = 0.1 A
Potential Difference (V) = 0.2 V
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
0.2 = 0.1 x R
⇒ R =
⇒ R = 2 Ω
Hence, resistance of filament of bulb = 2 Ω
Case 2 —
Current (I) = 400 m A = 0.4 A
Potentia Difference (V) = 1.0 V
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
1.0 = 0.4 x R
⇒ R =
⇒ R = 2.5 Ω
Hence, resistance of filament of bulb = 2.5 Ω
Therefore, we observe that with increase in temperature resistance of the wire increases.
Hence, resistance of filament increases with the increase in temperature.
Exercise 8A Short Question
22 questionsAnswer:
The potential difference (p.d.) between two points is equal to the work done per unit charge in moving a positive test charge from one point to the other. It's S.I. unit is volt. It is a scalar quantity.
Answer:
(a) Current is a scalar quantity.
By stating the direction of current in a conductor we mean that the direction of motion of electrons is opposite to it.
(b) Potential is a scalar quantity.
When the potential is positive at a point in the vicinity of a positive charge since work has to be done on the positive test charge against the repulsive force due to the positive charge in bringing it from infinity, while it is negative at a point in the vicinity of a negative charge since the attractive force on test charge due to the negative charge does work by itself.
(a) Name the particles which are responsible for the flow of current in a metallic wire.
(b) Explain the flow of current in a metallic wire on the basis of movement of the particles named by you above in part (a).
(c) What is the cause of resistance offered by the metallic wire in the flow of current through it?
Answer:
(a) The particles which are responsible for the flow of current in a metallic wire are free electrons.
(b) In a metallic wire, the moving charges are the free electrons which constitute the current.
If n electrons pass through the cross section of a conductor in time t, then total charge passed through the conductor is given as
Q (charge) = n × e
and the current in conductor is
(c) When the ends of a metal wire are connected to a cell, the electrons inside it experience a force in direction from the end at negative potential to positive, due to which they began to move as shown in the diagram below.

Due to force, their speed increases but during the movement they collide with the fixed positive ions and loose some of their kinetic energy due to which their speed decreases. This lost energy heats up the wire.
After the collision, they are again accelerated towards the positive potential due to the existing potential difference so their speeds again increases and then again in collision with the positive ions, their speed decreases. This process continues. As a result, the electrons do not move in bulk with a continuously increasing speed, but their is a drift of electrons towards the positive terminal. Thus, a metal wire offers some resistance to the flow of electrons through it.
Answer:
(a) The law which relates the potential difference and current in a conductor is known as Ohm's law.
Ohm's law states that the current flowing through the conductor is directly proportional to the potential difference across it's ends provided that the physical conditions and the temperature of the conductor remain constant.
(b) The necessary condition for a conductor to obey Ohm's law is that the physical conditions and the temperature should remain constant.
Answer:
Ohmic resistor | Non-ohmic resistor |
---|---|
It obeys the Ohm's law i.e., is constant for all values of V and I. | It does not obey the Ohm's law i.e., is not same for all values of V and I. |
The graph for potential difference V versus current I is a straight line passing through the origin. | The graph for potential difference V versus current I is not a straight line, but is a curve which may not pass through the origin. |
Answer:
Graph (a) is non-ohmic resistor and
Graph (b) is ohmic resistor.
The I-V graph for (b) is a straight line or linear while for (a), the graph is a curve.
Answer:
Resistance of a wire is directly proportional to the length of the wire.
R ∝ l
In a long conductor, the number of collisions of free electrons with the positive ions will be more as compared to a shorter one. Therefore, a longer conductor offers more resistance. Hence, the resistance of a conductor is directly proportional to the length of the wire.
Answer:
With the increase in temperature of a conductor, the random motion of electrons increases. As a result, the number of collisions of electrons with the positive ions increase. Hence, the resistance of a conductor increases with an increase in it's temperature.
Answer:
The iron wire will have more resistance.
Iron wire has more resistance as compared to copper because specific resistance of iron is more than that of copper.
Answer:
The three factors on which the resistance of wire depends are —
- Dependence on length of the wire — the resistance of a wire is directly proportional to the length of the wire.
R ∝ l - Dependence on the thickness of the wire — the resistance of a wire is inversely proportional to it's area of cross section (a) normal to the direction of flow of current.
R ∝ - Dependence on the temperature of the wire — the resistance of conductor increases with an increase in it's temperature.
Answer:
(a) Two factors on which the specific resistance of a wire depends are —
- Material of the substance — It is a characteristic property of the substance. It is different for different substances.
- Temperature of the substance — It increases with the increase in temperature for metals, but it decreases with the increase in temperature for the semiconductor.
(b) Both the wires will have the same specific resistance because the specific resistance is a characteristic property of the material and as both the wires are of copper hence both will have same specific resistance.
Answer:
(a) Dependency of resistance of a wire is as follows:
- Resistance of a wire is directly proportional to length of the wire.
- Resistance of a wire is inversely proportional to the square of radius of the wire.
(b) Dependency of specific resistance of a wire is as follows:
- Specific resistance of a wire does not depend on length of the wire as it is a characteristic property of the material of wire.
- Specific resistance of a wire does not depend on radius of the wire as it is a characteristic property of the material of wire.
Answer:
(a) Copper or aluminium materials are used for making connection wires because their specific resistance is very low and hence they possess least possible resistance.
Due to low (or negligible) resistance of connection wires, the current in circuit remains unaffected, and the loss of energy due to heating is prevented. Hence, they are made of materials such as copper or aluminium, whose specific resistance is very small.
(b) The connection wires are made thick so that their resistance becomes low.
Answer:
The material used for making fuse wire is an alloy of lead and tin because it's melting point is low and it's specific resistance is more than that of copper or aluminium so that the resistance of a short and thin fuse wire is more than that of the connecting wire. It permits current upto it's safe limit to pass through it, but an excessive current melts it so that it blows off and the circuit is broken.
Answer:
A superconductor is a substance of zero resistance (or infinite conductance) at a very low temperature.
Example — Mercury below 4.2 K, lead below 7.25 K and niobium below 9.2 K
Resistances of these substances decrease tremendously with the decrease in temperature and become almost zero in the low temperature range near absolute zero. Zero resistance of a superconductor means it's infinite conductivity (i.e., once a current starts flowing in a superconductor, it persists even when there is no potential across it).
Exercise 8A Very Short Question
5 questionsAnswer:
The expression is —
R = ρ
where
R = resistance of wire
ρ = specific resistance of the material of wire
l = length of wire
a = area of cross section of wire
Answer:
The order of specific resistance is as follows:
- The specific resistance is very low for metals, because it allows most of current to pass through it.
- The specific resistance is low for semiconductor.
- The specific resistance is very high for insulators, as the current won't pass through it.
Answer:
(i) The material used for the filament of an electric bulb is a tungsten wire because it has a high melting point.
(ii) The material used for the heating element of a room heater is nichrome because the specific resistance of nichrome is high and it's resistance increases to a great extent with the increase in temperature.
Exercise 8B Long Questions
4 questionsAnswer:
e.m.f — e.m.f of a cell is defined as the energy spent (or the work done) per unit charge in taking a positive test charge around the complete circuit of the cell. (i.e., in the circuit outside the cell as well as in the electrolyte inside the cell).
Terminal voltage — The terminal voltage of a cell is defined as the work done per unit charge in carrying a positive test charge around the circuit connected across the terminals of the cell.
Internal resistance — The resistance offered by the electrolyte inside the cell, to the flow of current, is known as the internal resistance of the cell.
Answer:
In a closed circuit, the current flows through the circuit. There is a fall of potential across the internal resistance of the cell. Hence, the p.d. across the terminals in a closed circuit is less than the p.d. across the terminals in an open circuit by an amount equal to the potential drop across the internal resistance of the cell.
Hence, the p.d. across the terminals of a cell is more in an open circuit and reduced in a closed circuit.
Answer:
Below diagram shows two resistors connected in series:

If current I is drawn from the battery, the current through each resistor will also be I.
By Ohm's law,
p.d. between A and B is V1 = VA - VB = IR1
p.d. between B and C is V2 = VB - VC = IR2
Adding these we get,
V = V1 + V2
= VA - VB + VB - VC
= VA - VC
= IR1 + IR2
= I (R1 + R2) [Equation 1]
If the equivalent resistance between the points A and C is RS, then the potential difference between the points A and C is
V = VA - VC = IRS [Equation 2]
Therefore from equation 1 and 2,
IRS = I (R1 + R2)
⇒ RS = R1 + R2
Thus, in the series combination, the equivalent resistance is equal to the sum of the individual resistances.
Answer:
Below diagram shows two resistors connected in parallel:

Let I1 and I2 be the currents through the resistances R1 and R2 respectively, then total current drawn from the battery is
I = I1 + I2 [Equation 1]
If potential difference between the two ends A and B is V, then by Ohm's law
current in R1 is I1 =
current in R2 is I2 =
On adding these,
I1 + I2 = + [Equation 2]
If the equivalent resistance of the combination between the points A and C is Rp, then total current drawn from the source is
I = [Equation 3]
Substituting the values of I and I1 + I2 from equation 3 and 2 in 1, we get,
Thus, in the parallel combination, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances.
Exercise 8B Multiple Choice Type
7 questionsAnswer:
the material of electrodes
Reason — The emf of a cell depends on :
- the material of electrodes and
- the electrolyte used in the cell.
Answer:
equal to, greater than
Reason — Emf of a cell is equal to the terminal voltage when cell is not in use and when the current starts to flow in the circuit, then there is a voltage drop due to internal resistance, thus we get, V = E - Ir. Hence, Emf is greater than the terminal voltage when cell is in use.
Answer:
current is same in each resistance
Reason — Current is same in each resistance in series combination of resistances. The current has a single path to flow, hence, same current passes through each resistor.
Answer:
p.d. is the same across each resistance
Reason — Potential difference is same across each resistor in parallel combination of resistances, which is equal to the potential difference across the terminals of the battery (or source).
For parallel combination of resistances, which of the statements are correct ?
- On I-V graph, the slope of line is more.
- The potential difference across each resistance is same.
- The current in a resistor is inversely proportional to the resistance.
- All of the above
Answer:
All of the above
Reason — For parallel combination of resistances :
- On I-V graph, the slope of line is more because resistance in parallel combination is less than series combination.
- The potential difference across each resistance is same because, the ends of each resistor are connected to the ends of the same source of potential.
- According to Ohm's Law V = IR. Hence, the current in a resistor is inversely proportional to the resistance.
Assertion (A): The terminal voltage of a cell is always less than its e.m.f.
Reason (R): More the current drawn from the cell, the less is the terminal voltage.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true and R is not the correct explanation of A
- assertion is false but reason is true.
- assertion is true but reason is false.
Answer:
assertion is false but reason is true..
Explanation
Assertion (A) is false. E.m.f. is greater than the terminal voltage only when the cell delivers the current to the external circuit.
Reason (R) is true. When a heavy current is drawn from the cell, a large number of charge carriers flow through the electrolyte and hence more work is done. This results in more voltage drop v, and hence less terminal voltage V. Thus, terminal voltage V of a cell depends on the amount of current I drawn from it.
Exercise 8B Numericals
31 questionsThe diagram in figure shows a cell of e.m.f. ε = 4 volt and internal resistance r = 2 ohm connected to an external resistance R = 8 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when (i) the key K is open, and (ii) the key K is closed.

Answer:
Given,
e.m.f. ε = 4 volt
resistance r = 2 ohm
external resistance R = 8 ohm
(i) When the key is open then no current is flowing in the circuit and hence the ammeter reading = 0
From relation,
Voltage (V) = ε – Ir
Substituting the values in the formula we get,
V = 4 – (0 × 2)
V = 4 volt
Hence, voltmeter reading = 4 volt
(ii) When key is closed, current drawn from the cell,
Substituting the values in the formula above we get,
I =
⇒ I =
⇒ I = 0.4 ampere
Hence, ammeter reading = 0.4 ampere
From relation,
Voltage (V) = ε – Ir
Substituting the values in the formula we get,
V = 4 - (0.4 x 2) = 3.2 V
Hence, voltmeter reading = 3.2 volt
Answer:
Given,
e.m.f. (ε) = 6.0 V
Current (I) = 3 A
Potential difference (V) = 5.4 V
r = ?
From relation,
V = ε – Ir
Substituting the values in the formula above we get,
5.4 = 6 - (3r)
⇒ r = 6 - 5.4
⇒ 3r = 0.6
⇒ r =
⇒ r =
⇒ r =
⇒ r = 0.2 Ω
Hence, internal resistance of the battery = 0.2 Ω
Answer:
(a) Given,
e.m.f. (ε) = 1.8 V
Internal resistance (r) = 2 Ω
I = ?
Total resistance of arrangement = 2 + 0.7 + 4.5 = 7.2 Ω
From relation,
Substituting the values in the formula above we get,
I =
⇒ I = 0.25 A
Hence, reading of ammeter = 0.25 A
(b) Current (I) = 0.25 A
total resistance (excluding internal resistance) = 4.5 + 0.7 = 5.2 ohm
Using ohm's law
V = IR
Substituting the values in the formula above we get,
V = 0.25 × 5.2
V = 1.3 V
Hence, potential difference across the terminals of the battery = 1.3 V
A music system draws a current of 400 mA when connected to a 12 V battery.
(a) What is the resistance of the music system ?
(b) The music system if left playing for several hours and finally the battery voltage drops and the music system stops playing when the current drops to 320 mA. At what battery voltage does the music system stop playing.
Answer:
(a) Given,
I = 400 mA = 400 x 10-3 = 0.4 A
V = 12 V
From Ohm's law,
V = IR
Substituting the values in the formula we get,
12 = 0.4 x R
⇒ R =
⇒ R = 30 Ω
Hence, the resistance of the music system = 30 Ω
(b) Given,
I = 320 mA = 320 x 10-3 A = 0.32 A
R = 30 Ω
From Ohm's law,
V = IR
Substituting the values in the formula we get,
V = 0.32 x 30 = 9.6V
Hence, the battery voltage when the music system stops playing = 9.6 V
Answer:
Given,
e.m.f. = ε
internal resistance = r
current (I) = 1.0 A
external resistance (R) = 1.9 ohm
Case 1
From relation,
ε = I (R + r)
Substituting the value in the formula above we get,
ε = 1 (1.9 + r)
ε = 1.9 + r [Equation 1]
In second case,
I = 0.5 A,
R = 3.9 Ω
Substituting the value in the formula above we get,
ε = 0.5 (3.9 + r)
ε = 1.95 + 0.5r [Equation 2]
Equating 1 and 2 we get,
Now, substituting the value of r in equation 1, we get,
ε = 1.9 + 0.1
ε = 2 V
Hence, ε = 2 V , r = 0.1 Ω
Answer:
In order to get a total resistance less than 1 Ω, the three resistors of values 2 Ω, 3 Ω and 5 Ω, should be connected in parallel as shown in the diagram below:

In parallel, if the equivalent resistance is RP, then
Hence, the three resistors should be connected in parallel and total resistance = 0.97 Ω
Answer:
When two resistors (2 Ω each) in parallel combination are connected to the third resistor (2 Ω) in series connection then the resultant resistance is 3 Ω. The below diagram shows this arrangement of resistors:

To verify,
Total resistance of two resistors in parallel,
In parallel, if the equivalent resistance is RP, then
Hence, RP = 1 Ω
In series connection,
Equivalent resistance = RE = RP + R3
= 1 + 2 = 3 Ω
Hence, Equivalent resistance = 3 Ω
Answer:
Let the resistors be R1, R2, R3, R4 each of 2.0 Ω.
In parallel, if the equivalent resistance is RP, then
Hence, RP = 1 Ω
In series connection,
Equivalent resistance = RE = R1 + R2 + RP
= 2 + 2 + 1 = 5 Ω
Hence, Equivalent resistance = 5 Ω
Answer:
Let the four set of resistors be R1, R2, R3, R4 .
As each set consists of three resistors of 2 Ω each in series:
Resistance of each set = 2 + 2 + 2 = 6 Ω
As four similar sets of equivalent resistance 6 Ω are arranged in parallel, so the equivalent resistance (RP) will be:
Hence, equivalent resistance = 1.5 Ω
Answer:
In the circuit, there are three parts. In the first part, resistors of 4 Ω and 8 Ω are connected in series. If the equivalent resistance of this part is R's then
R's = 4 + 8 = 12 Ω
In the second part, resistors of x Ω and 5 Ω are connected in series. If the equivalent resistance of this part is R''s then
R''s = (x + 5) Ω
In the third part, the two parts of resistance R's = 12 Ω and R''s = (x + 5) Ω are connected in parallel. If the equivalent resistance between points A and B is Rp then
But it is given that equivalent resistance between points A and B is 4 Ω
∴ Rp = 4 Ω
Putting value of Rp in the equation above and solving for x:
Hence, the value of x = 1 Ω
Answer:
In the circuit, there are four parts. In the first part, three resistors of 1 Ω each are connected in series. If the equivalent resistance of this part is R's then
R's = (1 + 1 + 1) Ω = 3 Ω
In the second part, three resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R''s then
R''s = (2 + 2 + 2) Ω = 6 Ω
In the third part, the two parts of resistance R's = 3 Ω and R''s = 6 Ω and 2 Ω are connected in parallel. If the equivalent resistance is Rp then
∴ Rp = 1 Ω
In the fourth part 1 Ω, (Rp = 1 Ω) and 1 Ω are connected in series in between points A and B.
Hence, the equivalent resistance is 1 + 1 + 1 = 3 Ω
Answer:
Given, uniform wire with a resistance of 27 Ω is divided into three equal pieces. Hence, resistance of each piece = 9 Ω.
Three such pieces are joined in parallel.
In parallel, if the equivalent resistance is RP, then
Hence, equivalent resistance = 3 Ω
Answer:
Below diagram shows the arrangement of resistors:

In the circuit, there are two parts. In the first part, two resistors of 6 Ω and 12 Ω are connected in parallel. If the equivalent resistance is Rp then
Hence, RP = 4 Ω
In the second part, 1 Ω and (RP = 4 Ω) are connected in series. If the equivalent resistance is Rs then
Rs = 1 + RP = 1 + 4 = 5 Ω
Hence, total resistance of the circuit = 5 Ω
Answer:
In the circuit, there are two parts. In the first part, resistors of 12 Ω, 6 Ω and 4 Ω are connected in parallel. If the equivalent resistance of this part is R'p then
Hence, R'P = 2 Ω
In the second part, resistors of 2 Ω, (R'p = 2Ω) and 5 Ω are connected in series. If the equivalent resistance of this part is Rs then
Rs = (2 + 2 + 5) Ω = 9 Ω
Hence, Effective resistance between A and B = 9 Ω
Answer:
In the circuit, there are three parts. In the first part, two resistors of 3 Ω, 2 Ω are connected in series. If the equivalent resistance is R's then
R's = 3 + 2 = 5 Ω
In the second part, two resistors of 6 Ω, 4 Ω are connected in series. If the equivalent resistance is R''s then
R''s = 6 + 4 = 10 Ω
In the third part R's, 30 Ω and R''s are connected in parallel. If the equivalent resistance is Rp then
Hence, RP = 3 Ω
Answer:
In the circuit, there are two parts. In the first part, three resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R's then
R's = (2 + 2 + 2) Ω = 6 Ω
In the second part, the resistance of first part (R's = 6 Ω ) and 2 Ω are connected in parallel. If the equivalent resistance of this part is Rp then
∴ Rp = 1.5 Ω
(b) In the circuit, there are three parts. In the first part, two resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R's then
R's = (2 + 2) Ω = 4 Ω
In the second part, two resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R''s then
R''s = (2 + 2) Ω = 4 Ω
In the third part, the two parts of resistance R's = 4 Ω and R''s = 4 Ω are connected in parallel. If the equivalent resistance between points A and C is Rp then
∴ Equivalent Resistance between A and C = 2 Ω
Answer:
(a) In the circuit, there are two parts. In the first part, two resistors of 3 Ω each are connected in series. If the equivalent resistance of this part is R's then
R's = (3 + 3) Ω = 6 Ω
In the second part, resistance R's = 6 Ω and 3 Ω are connected in parallel. If the equivalent resistance between points P and Q is Rp then
∴ Equivalent resistance between the points P and Q = 2 Ω
(b) In the circuit, 3 Ω, Rp = 2 Ω and 3 Ω are connected in series. If the equivalent resistance of this part is Rs then
Rs = (3 + 2 + 3) Ω = 8 Ω
∴ Equivalent resistance between the points X and Y = 8 Ω
Answer:
(a) Circuit diagram showing two resistors of 4.0 Ω and 6.0 Ω connected in series with a battery of 6.0 V and negligible internal resistance is shown below:

Given,
Two resistors of 4 Ω and 6 Ω are connected in series. If the equivalent resistance of this part is R's then
R's = (4 + 6) Ω = 10 Ω
Potential Difference V = 6 V
Current I = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
6 = I x 10
⇒ I = 6 / 10 = 0.6 A
Hence, in series, current through the battery = 0.6 A
(b) Circuit diagram showing two resistors of 4.0 Ω and 6.0 Ω connected in parallel with a battery of 6.0 V and negligible internal resistance is shown below:

Given,
Two resistors of 4 Ω and 6 Ω are connected in parallel. If the equivalent resistance of this part is Rp then
Potential Difference V = 6 V
Current I = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
6 = I x 2.4 ⇒ I = 6 / 2.4 = 2.5 A
Hence, in parallel, current through the battery = 2.5 A
Answer:
(a) Given,
Two resistors of 6 Ω and 4 Ω are connected in series. If the equivalent resistance of this part is R's then
R's = (6 + 4) Ω = 10 Ω
Potential Difference V = 20 V
Current I = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
20 = I x 10
⇒ I = 20 / 10 = 2 A
Hence, in series, current through the battery = 2 A
(b) Given,
Resistance R = 6 Ω
Potential Difference V = ?
Current I = 2 A
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
V = 2 x 6 = 12 V
Hence, potential difference = 12 V
Answer:
(a) Labelled diagram of the arrangement is shown below:

(b) Two resistors of 2 Ω and 3 Ω are connected in parallel. If the equivalent resistance of this part is Rp then
Hence, equivalent resistance = 1.2 Ω
Current I = 0.5 A
Potential Difference V = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
V = 0.5 x 1.2 = 0.6 V
Current through 2 Ω = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
0.6 = I x 2 ⇒ I = 0.6 / 2 = 0.3 A
Hence, current through 2 Ω resistor = 0.3 A
Current through 3 Ω = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
0.6 = I x 3 ⇒ I = 0.6 / 3 = 0.2 A
Hence, current through 3 Ω resistor = 0.2 A
Answer:
(a) Given,
R = 1 Ω
V = 2 V
Current through 1 Ω = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
2 = I x 1
⇒ I = 2 / 1 = 2 A
Hence, current through Resistor A = 2 A
(b) Given,
R = 2 Ω
V = 2 V
Current through 2 Ω = ?
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
2 = I x 2
⇒ I = 2 / 2 = 1 A
Hence, current through Resistor B = 1 A
Answer:
(a) Given,
V = 4 V
I = 0.4 A
From Ohm's law
V = IR
4 = 0.4 x R
⇒ R = 4 / 0.4 = 10 Ω
Hence, total resistance of the circuit = 10 Ω
(b) As total resistance of the circuit is equal to 10 Ω and the two resistors R and 20 Ω are connected in parallel, hence we get,
Hence, the value of R = 20 Ω
(c) current flowing in R = ?
V = 4 V
R = 20 Ω
I = ?
From Ohm's law
V = IR
⇒ 4 = I x 20
⇒ I = 4 / 20 = 1 / 5 = 0.2 A
Hence, the current flowing in R = 0.2 A
A particular resistance wire has a resistance of 3.0 ohm per meter. Find —
(a) The total resistance of three lengths of this wire each 1.5 m long, joined in parallel.
(b) The potential difference of the battery which gives a current of 2.0 A in each of the 1.5 m length when connected in parallel to the battery (assume that resistance of the battery is negligible).
(c) The resistance of 5 m length of a wire of the same material, but with twice the area of cross section.
Answer:
(a) Resistance of 1 m of wire = 3 Ω.
Resistance of 1.5 m of wire = 3 x 1.5 = 4.5 Ω. As three such wires are joined in parallel and if the equivalent resistance of this part is Rp then
Hence, total resistance of circuit = 1.5 Ω
(b) I = 2 A
From Ohm's Law
V = IR
Substituting the values in the formula above we get,
V = 2 x 4.5 = 9 V
Hence, potential difference = 9 V
(c) R = 3 Ω for 1 meter wire
Therefore, for 5 m
R = 3 x 5 = 15 Ω
Here the area is twice and resistance is inversely proportional to area.
Thus, resistance becomes half
R = 15 / 2 = 7.5 Ω
Hence, resistance = 7.5 Ω
Answer:
(i) Given,
Two resistors each of 2 Ω connected in parallel. If the equivalent resistance of this part is Rp then
Hence, equivalent resistance Rp = 1 Ω
Given,
I = 1.2 A
From relation,
ε = I (R + r)
ε = 1.2 (1 + r)
ε = 1.2 + 1.2r [Equation 1]
When the resistors are connected in series, it supplies a current of 0.4 A,
If the equivalent resistance of this part is Rs then
Rs = 2 + 2 = 4 Ω
Hence, equivalent resistance Rs = 4 Ω
Given, I = 0.4 A
From relation,
ε = I (R + r)
ε = 0.4 (4 + r)
ε = 1.6 + 0.4r [Equation 2]
Equating 1 and 2, we get,
1.2 + 1.2r = 1.6 + 0.4r
⇒ 1.2r – 0.4r = 1.6 - 1.2
⇒ 0.8r = 0.4
⇒ r =
⇒ r = 0.5 Ω
Hence, Internal resistance r = 0.5 Ω
(ii) Substituting the value in equation 1 we get,
ε = 1.2(1 + r)
= 1.2 (1 + 0.5)
= 1.2 x 1.5
= 1.8 V
Hence, e.m.f. of the cell = 1.8 V
A battery of e.m.f. 16 V and internal resistance 2 Ω is connected to two resistors 3 Ω and 6 Ω connected in parallel. Find (a) the current through the battery (b) p.d. between the terminals of the battery (c) the current in 3 Ω resistor (d) the current in 6 Ω resistor.
Answer:
(a) Given,
e.m.f. = 16 V
internal resistance r = 2 Ω
current through battery = ?
If Rp is the equivalent resistance of resistors 3 Ω and 6 Ω connected in parallel, then
From relation,
ε = I (R + r)
Substituting the value in the formula above we get,
16 = I(2 + 2)
⇒ 16 = I x 4 ⇒ I = 16 / 4 = 4 A
Hence, current through the battery = 4 A
(b) Potential difference between the terminals of the battery = ?
Using Ohm's law
V = IR
R = 2 Ω
I = 4 A
Substituting the values in the formula above we get,
V = 4 x 2 = 8 V
Hence, potential difference between the terminals of the battery = 8 V
(c) Current in 3 Ω resistor = ?
Using Ohm's law
V = IR
R = 3 Ω
V = 8 V
I = ?
Substituting the values in the formula above we get,
8 = I × 3
I =
⇒ I = 2.66 A
Hence, current in 3 Ω resistor is 2.66 A
(d) Current in 6 Ω resistor = ?
Using Ohm's law
V = IR
R = 6 Ω
V = 8 V
I = ?
Substituting the values in the formula above we get,
8 = I × 6
⇒ I = 8 / 6 = 1.333 A
Hence, current in 6 Ω resistor is 1.34 A
The circuit diagram in figure shows three resistors 2 Ω, 4 Ω and R Ω connected to a battery of e.m.f. 2V and internal resistance 3 Ω. If main current of 0.25 A flows through the circuit, find —
(a) the p.d. across the 4 Ω resistor
(b) the p.d. across the internal resistance of the cell,
(c) the p.d. across the R Ω or 2 Ω resistor, and
(d) the value of R.

Answer:
(a) Given,
resistor = 4 Ω
I = 0.25 A
the p.d. across the 4 Ω resistor (V) = ?
Using Ohm's law
V = IR
Substituting the values in the formula above we get,
V = 0.25 x 4 = 1 V
Hence, the p.d. across the 4 Ω resistor (V) = 1 V
(b) Given,
internal resistance = 3 Ω
I = 0.25 A
the p.d. across the internal resistance of the cell = ?
Using Ohm's law
V = IR
Substituting the values in the formula above we get,
V = 0.25 x 3 = 0.75 V
Hence, the p.d. across the internal resistance (V) = 0.75 V
(c) Potential difference across R Ω or 2 Ω
V = Vnet - Vacross 4 Ω - Vacross 3 Ω
Hence, we get,
V = 2 - 1 - 0.75 = 0.25 V
(d) The p.d. across resistor of R Ω = 0.25 V
Let the equivalent resistance of the resistors of 2 Ω and R Ω connected in parallel be R'p
Using Ohm's law
V = IR
0.25 = 0.25 x R'p
Substituting the value of R'p from above:
0.25 = 0.25 x
⇒ = 1
⇒ 2R = 2 + R
⇒ 2R - R = 2
⇒ R = 2 Ω
Hence, value of R = 2 Ω
Answer:
(a) In the circuit, there are two parts. In the first part, resistors of 2.0 and 4.0 Ω are connected in series. If the equivalent resistance of this part is Rs then
Rs = 2 + 4 = 6 Ω
In the second part, Rs = 6.0 and resistor of 6.0 Ω are connected in parallel. If the equivalent resistance of this part is Rp then
Hence, the effective resistance of the circuit = 3 Ω
(b) The reading of ammeter = ?
R = 3 Ω
V = 6.0 V
Using Ohm's law,
V = IR
Substituting the values in the formula above we get,
6 = I x 3
⇒ I = 6 / 3 = 2 A
Hence, the reading of ammeter = 2 A
Answer:
(a) In the circuit, there are three parts. In the first part, resistors of 10 Ω and 40 Ω are connected in parallel. If the equivalent resistance of this part is R'p then
In the second part, resistors of 30 Ω, 20 Ω and 60 Ω are connected in parallel. If the equivalent resistance of this part is R''p then
In the third part, resistors R'p and R''p are connected in series. If the equivalent resistance of this part is Rs then
Hence, the total resistance of the circuit = 18 Ω
(b) Given,
e.m.f. = 1.8 V
effective resistance of the circuit = 18 Ω
I = ?
From Ohm's law
V= IR
Substituting the values in the formula above, we get,
1.8 = I x 18
⇒ I =
⇒ I = 0.1 A
Hence, the reading of ammeter = 0.1 A
A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in figure.

Find —
(a) the reading of the ammeter,
(b) the potential difference across the terminals of the cell, and
(c) the potential difference across the 4.5 Ω resistor.
Answer:
(a) Given,
e.m.f. = 2V
I = ?
In the circuit, there are two parts. In the first part, resistors of 4.5 Ω and 9 Ω are connected in parallel. If the equivalent resistance of this part is Rp then
In the second part, 1.2 Ω, 0.8 Ω and Rp = 3 Ω are connected in series. If the equivalent resistance of this part is Rs then
Hence, the effective resistance of the circuit = 5 Ω
Using Ohm's law,
V = IR
2 = I x 5
⇒ I =
⇒ I = 0.4 A
Hence, the reading of the ammeter = 0.4 A
(b) The potential difference across the ends of the cells = ?
ε = 2 V
I = 0.4 A
r = 1.2 Ω
From relation,
Voltage (V) = ε – Ir
Substituting the values in the formula we get,
V = 2 - (0.4 x 1.2)
⇒ V = 2 - 0.48
⇒ V = 1.52 V
Hence, potential difference across the terminals of the cell = 1.52 V
(c) The potential difference across the 4.5 Ω resistor = ?
Current flowing is I = 0.4 A. Now the current I divides in two parts. Let the current in 4.5 Ω resistor be I1 and in 9 Ω resistor be I2.
So I = I1 + I2
and I1 x 4.5 = I2 x 9
On solving,
p.d. across the 4.5 Ω resistor
Alternate Method:
V4.5Ω = Vcell - Vammeter
Hence, p.d. across the 4.5 Ω resistor = 1.2 V
Exercise 8B Short Questions
5 questionsAnswer:
E.m.f of cell | Terminal voltage of cell |
---|---|
It is measured by the amount of work done per unit change in moving a positive test charge in the complete circuit inside and outside the cell. | It is measured by the amount of work done per unit charge in moving a positive test charge in the circuit outside the cell. |
It is the the characteristic of the cell, i.e., it does not depend on the amount of current drawn from the cell. | It depends on the amount of current drawn from the cell. More the current drawn from the cell, less is the terminal voltage. |
Answer:
The factors on which the internal resistance of a cell depends are —
- The surface area of the electrodes — larger the surface area of electrodes, less is the internal resistance.
- The distance between the electrodes — more the distance between the electrodes, greater is the internal resistance.
Answer:
Given,
e.m.f. = ε
internal resistance = r
external resistance = R
(a) The total resistance of circuit = R + r
(b) The current drawn from the cell
I = =
(c) the p.d. across the cell = V = IR
Substituting from above,
V = x R
(d) voltage drop inside the cell V = Ir
Substituting from above,
V = x r
Answer:
(a) When a cell is used to send current to an external circuit, it's terminal voltage V is less than it's e.m.f.
Hence, terminal voltage < e.m.f.
(b) The emf of the cell is equal to it's terminal voltage when no current is drawn.
Answer:
The slope of V-I graph gives the resistance. Since the straight line A is less steeper than B, so the straight line A represents small resistance. In parallel combination, the equivalent resistance is less than in series combination so A represents the parallel combination.
Exercise 8B Very Short Questions
3 questionsAnswer:
(a) Equivalent resistance R of three resistors R1, R2 and R3 joined in parallel is given by —
= + +
(b) Equivalent resistance R of three resistors R1, R2 and R3 joined in series is given by —
R = R1 + R2 + R3
State how are the two resistors joined with a battery in each of the following cases when —
(a) same current flows in each resistor,
(b) potential difference is same across each resistor,
(c) equivalent resistance is less than either of the two resistances,
(d) equivalent resistance is more than either of the two resistances.
Answer:
(a) When same current flows in each resistor, then the two resistors are joined in series
(b) When potential difference is same across each resistor, then the two resistors are joined in parallel
(c) When equivalent resistance is less than either of the two resistances, then the two resistors are joined in parallel
(d) When equivalent resistance is more than either of the two resistances, then the two resistors are joined in series
Answer:
(a) In the circuit, two resistors of 2 Ω each are connected in parallel. In parallel, the equivalent resistance is RP, then
Hence, RP = 1 Ω
(b) In the circuit, three resistors of 1 Ω, 1 Ω and 2 Ω are connected in parallel. In parallel, the equivalent resistance is RP, then
Hence, RP = 0.4 Ω
(c) In the circuit two resistors of 1 Ω each are connected in parallel. In parallel, the equivalent resistance is RP, then
Hence, RP = 0.5 Ω
(d) In the circuit there are three parts. In the first part two resistor of 1 Ω each are connected in series. In series, the equivalent resistance is R'S, then
R'S = (1 + 1) Ω = 2 Ω
In the second part two resistor of 1 Ω each are connected in series. In series, the equivalent resistance is R''S, then
R''S = (1 + 1) Ω = 2 Ω
In the third part,R'S and R''S are in parallel, the equivalent resistance is RP, then
Hence, RP = 1 Ω
Therefore, on observing the values we can say that (a) and (d) have same equivalent resistances i.e., 1 Ω.
Exercise 8C Long Questions
3 questionsAnswer:
Appliance | Power (in kilowatt) | Voltage (V) | Time (t) | Electrical energy (E = P × t) in (kWh) |
---|---|---|---|---|
Fluorescent tube | 0.04 | 220 | 6 | 0.24 |
Television set | 0.12 | 220 | 3 | 0.36 |
Room heater | 1 | 220 | 2 | 2 |
(i) Energy consumed by fluorescent tube in one day = 0.24 kWh
Hence, energy consumed in 30 days = 0.24 x 30 = 7.2 kWh.
(ii) Energy consumed by television set in one day = 0.36 kWh
Hence, energy consumed in 30 days = 0.36 x 30 = 10.8 kWh.
(iii) Energy consumed by room heater in one day = 2 kWh
Hence, energy consumed in 30 days = 2 x 30 = 60 kWh.
Answer:
Resistance of lamp A (rating 220 V, 50 W) is —
Substituting the values in the formula above we get,
Resistance of lamp B (rating 220 V, 100 W) is —
Substituting the values in the formula above we get,
From formula, we get, P = I2 R
As we observe that RA > RB and same current is flowing in both the lamps as both are connected in series therefore PA > PB. Hence, 50 W lamp consumes more power than 100 W lamp.
Answer:
The amount of heat produced in a wire on passing current through it, depends on the following three factors.
- The amount of current passing through the wire — The amount of heat H produced in a wire is directly proportional to the square of current I passing through the wire, i.e., H ∝ I2
- The resistance of wire — The amount of heat H produced in the wire is directly proportional to the resistance R of the wire, i.e., H ∝ R.
- The time for which current is passed in the wire — The amount of heat H produced in a wire is directly proportional to the time t for which current is passed in the wire i.e., H ∝ t
Exercise 8C Multiple Choice Type
5 questionsAnswer:
Both (1) and (3)
Reason — The amount of heat produced in a wire on passing current through it, depends on the following three factors.
- The amount of current passing through the wire — The amount of heat H produced in a wire is directly proportional to the square of current I passing through the wire, i.e., H ∝ I2
- The resistance of wire — The amount of heat H produced in the wire is directly proportional to the resistance R of the wire, i.e., H ∝ R.
- The time for which current is passed in the wire — The amount of heat H produced in a wire is directly proportional to the time t for which current is passed in the wire i.e., H ∝ t
Answer:
144 Ω
Reason — Resistance of an electrical appliance (rating 100 W, 120 V) is :
Substituting the values in the formula above we get,
Hence, resistance of an electrical appliance = 144 Ω
Answer:
nearly 25 W
Reason — Given,
Power (P) = 100 W
Voltage (V) = 220 V
We know that,
Power (P) = VI
Substituting the values in the formula above, we get,
100 = 220 x I
⇒ I =
⇒ I = 0.45 A
Hence, current through the lamp = 0.45 A
Power consumed when voltage is 110 V = ?
Resistance of lamp (R) =
Substituting the values in the formula, we get,
From relation,
P =
Substituting the values in the formula, we get,
Hence, power consumed is 25 W
Exercise 8C Numericals
21 questionsAnswer:
(a) Given,
Resistance (R) = 500 Ω
Current (I) = 0.4 A
Power = ?
From Ohm's law:
V = IR
Substituting the values in the formula above, we get,
V = 0.4 x 500 = 200 V
Power (P) = VI
Substituting the values in the power formula we get,
P = 200 x 0.4 = 80 W
Hence, the power of bulb = 80 W
(b) The potential difference at it's end = 200 V
Answer:
(a) Given,
Current (I) = 3 A
Resistance (R) = 75 Ω
Time (t) = 2 min = 120 s
Heat produced = ?
Using
H = I2Rt
Substituting the values in the formula above, we get,
H = 32 x 75 x 120
⇒ H = 81000 J
Hence, heat energy produced = 81000 J
(b) Charge passed through the resistance = ?
From relation,
Q = It
Substituting the values in the formula above we get,
Q = 3 x 120
⇒ Q = 360 C
Hence, charge passed through the resistance = 360 C
Answer:
Given,
Power (P) = 60 W
Voltage (V) = 250 V
We know that,
Power (P) = VI
Substituting the values in the formula above, we get,
60 = 250 x I
⇒ I =
⇒ I = 0.24 A
Hence, current through the lamp = 0.24 A
Power consumed when voltage is 200 V = ?
Resistance of lamp (R) =
Substituting the values in the formula, we get,
From relation,
P =
Substituting the values in the formula, we get,
Hence, power consumed reduces to 38.4 W
Answer:
Given,
Power (P) = 100 W
Voltage (V) = 250 V
From relation,
Power (P) = VI
Substituting the values in the formula above, we get,
100 = 250 x I
⇒ I =
⇒ I = 0.4 A
Hence, current drawn = 0.4 A
Answer:
(a) Given,
P = 100 W
V = 220 volt
We know that,
Power P =
Substituting the values in the formula above, we get,
100 =
⇒ R =
⇒ R = 484 Ω
Hence, resistance of electric bulb = 484 Ω
(b) From relation P = VI
Safe current I =
Substituting the value we get,
I =
⇒ I = 0.45 A
Hence, safe current limit = = 0.45 A
Answer:
Given,
Power (P) = 60 W
time (t) = 12.5 h for 30 days
As energy consumed (E) = P × t
E = 60 x 12.5
⇒ E = 750 Wh
Hence, electrical energy consumed in one day = 750 Wh
Electrical energy consumed in 30 days = ?
E = 750 x 30
E = 22500 Wh
E = 22.5 kWh
Hence, electrical energy consumed in 30 days = 22.5 kWh
Answer:
Given,
Power (P) = 750 W
time (t) = 16 h
As energy consumed (E) = P × t
E = 750 x 16
⇒ E = 12000 Wh
⇒ E = 12 kWh
Hence, electrical energy consumed in 16 hours = 12 kWh
Answer:
(i) Given,
Resistance (R) = 200 Ω
Voltage (V) = 200 volt
Time (t) = 5 min = 300 sec
As we know,
Energy (E) =
Substituting the values in the formula above we get,
E =
⇒ E = 60,000 J
Hence, energy consumed = 60,000 J
(ii) In kWh = ?
As we know,
1 kWh = 3.6 x 106 J
1 J = kWh
60,000 J =
⇒ 60,000 J= 0.0167 kWh
Hence, energy consumed = 0.0167 kWh
Answer:
(i) Given,
Voltage (V) = 12 V
Power (P) = 24 W
Current (I) = ?
From relation,
Power (P) = VI
Substituting the values in the formula above, we get,
24 = 12 x I
⇒ I =
⇒ I = 2 A
Hence, the current flowing through it = 2 A
(ii) Energy (E) = P × t
Time (t) = 20 min = 1200 sec
E = 24 x 1200 = 28,800 J
Hence, energy consumed = 28,800 J
Answer:
(i) Given,
Current (I) = 0.2 A
Potential difference (V) = 20 V
From Ohm's law
V = IR
Substituting the values in the formula above, we get,
20 = 0.2 x R
⇒ R =
⇒ R =
⇒ R = 100 Ω
Hence, the resistance of the wire = 100 Ω
(ii) Given,
Time (t) = 60 sec
heat energy produced in 1 minute = ?
From relation,
Heat energy (H) = I2Rt
Substituting the values in the formula above, we get,
H = 0.22 x 100 x 60 = 240 J
Answer:
Given,
Voltage (V) = 240 V
Power (P) = 60 W
Resistance (R) = ?
From relation
We get,
Hence, resistance of the electric lamp = 960 Ω
From the relation for current through the element, we know,
Substituting the values in the formula above, we get,
When one lamp is connected across the mains, it draws 0.25 A current. If two such lamps are connected in series across the mains, current through each bulb becomes —
i.e., current is halved. Hence, heating (= I2Rt) in each bulb becomes one fourth, so each bulb appears less bright.
Answer:
Given,
Voltage (V1) = 220 V
Voltage (V2) = 110 V
P = 60 W (both bulbs have same power)
Resistance (R) = ?
From relation
Case 1:
Case 2:
Ratio between the two is —
Hence, ratio of resistances of the two bulbs is = 4 : 1
Answer:
(i) Given,
Power (P) = 250 W
Voltage (V) = 230 V
Time (t) = 1 h = 3600 sec
Energy (E) = ?
From relation,
E = P × t
Substituting the values in the formula above, we get,
E = 250 x 3600 = 9 x 105 J
Hence, energy consumed in one hour = 9 x 105 J
(ii) Let t be the time in which the bulb will consume 1.0 kWh energy when connected to 230 V mains
P = 1.0 kWh = 1000 Wh
Substituting the value in the formula, we get,
1000 = 250 x t
⇒ t = 1000 / 250 = 4 h
Hence, time = 4 h
Answer:
(i) Given,
Power (P) = 250 W
Voltage (V) = 100 V
Current through each heater (I) = ?
As P = VI
Substituting the values in the formula above we get,
250 = 100 x I
⇒ I = 250 / 100 = 2.5 A
Hence, current through each heater = 2.5 A
Therefore, current for three heaters =
I = 3 x 2.5 = 7.5 A
Hence, total current taken from supply = 7.5 A
(ii) Resistance for each heater (R) = ?
Since, V = IR
Substituting the values in the formula above we get,
100 = 2.5 x R
⇒ R = 100 / 2.5 = 40 Ω
Hence, resistance of each heater = 40 Ω
(iii) Energy supplied to three heater in 5 h = ?
Energy (E) = P × t
Substituting the values in the formula above, we get,
E = 250 x 5 = 1250 Wh = 1.25 kWh
Hence, Energy for three heaters = 3 × 1.25 = 3.75 kWh
Hence, energy supplied to the three heaters = 3.75 kWh
A bulb is connected to a battery of p.d. 4 V and internal resistance 2.5 ohm. A steady current of 0.5 A flows through the circuit. Calculate —
(i) the total energy supplied by the battery in 10 minutes,
(ii) the resistance of the bulb, and
(iii) the energy dissipated in the bulb in 10 minutes.
Answer:
(i) Given,
Voltage (V) = 4 V
Internal resistance = 2.5 Ω
Current (I) = 0.5 A
time (t) = 10 min = 600 sec
Energy supplied by the battery (E) = V x I x t
Substituting the values in the formula, we get,
E = 4 x 0.5 x 600 = 1200 J
Hence, total energy supplied = 1200 J
(ii) resistance of the bulb = ?
From relation V = I (R + r)
We get,
4 = 0.5 (R + 2.5)
⇒ 4 = (0.5 R) + (0.5 x 2.5)
⇒ 4 = (0.5 R) + 1.25
⇒ 0.5 R = 4 - 1.25
⇒ R = 2.75 / 0.5 = 5.5 Ω
Hence, resistance of the bulb = 5.5 Ω
(iii) Energy dissipated in the bulb in 10 min = ?
From relation
E = I2Rt
Substituting the values in the formula above, we get,
E = 0.5 x 0.5 x 5.5 x 600 = 825 J
Hence, energy dissipated = 825 J
Answer:
(i) Given,
Resistance, RA = 4 Ω
Resistance, RB = 6 Ω
Let equivalent resistance of the two resistors connected in parallel be RP
Hence, equivalent resistance = 2.4 Ω
From relation
We get,
Hence, power supplied = 15 W
(ii) Power dissipation across each resistor = ?
From relation, P = VI
Current across A (4 Ω) resistor —
IA = 6 / 4 = 1.5 A
Power dissipation across A = VIA
Substituting we get,
P = 6 x 1.5 = 9 W
Similarly,
Current across B ( 6 Ω) resistor —
IB = 6 / 6 = 1 A
Power dissipation across B resistor = VIB
Substituting the values in the formula above, we get,
P = 6 x 1 = 6 W
Hence, power dissipated across A = 9 W and across B = 6 W
Answer:
(i) Given,
e.m.f. (V) = 15 V
Internal resistance = 2 Ω
Resistors are R1 = 4 Ω and R2 = 6 Ω
Electrical energy = ?
As the battery and resistors are connected in series, equivalent resistance is given by
R = 2 + 4 + 6 = 12 Ω
From Ohm's law —
V = IR
Substituting the values in the formula above, we get,
15 = I x 12
⇒ I = 15 / 12 = 1.25 A
Hence, current in circuit = 1.25 A
Now,
Voltage across 6 Ω = IR = 1.25 × 6 = 7.50 V
Hence, Voltage across 6 Ω = 7.5 V
Time (t) = 1 min = 60 sec
We know, E =
Substituting the values in the formula above, we get,
E =
⇒ E = 7.52 x 10
⇒ E = 562.5 J
Hence, energy spent across the 6 Ω resistor = 562.5 J
Answer:
Given,
V = 220 V
time (t) = 5 min = 300 sec
From relation
and
Heat produced (H) —
H = P x t
⇒ H = x t
Case 1 for V = 220 V
[Equation 1]
Case 2 for V = 200 V
[Equation 2]
Equating 1 and 2 we get,
⇒ t = 363 sec = 6.05 min
Hence, time taken = 6.05 min
Answer:
Given,
I = 8 A
V = 220 V
t = 2 h
rate = 4.50 per kWh
Cost = ?
From relation,
E = V x I x t
we get,
E = 220 x 8 x 2 = 3520 Wh = 3.52 kWH
As cost of energy for 1 kWh = ₹ 4.50
Therefore, cost of energy for 3.52 kWh = 4.5 x 3.52
= ₹15.84
Hence, cost of electrical energy = ₹15.84
Answer:
Given,
Power (P) = 2 kW
Voltage (V) = 220 V
Time (t) = 7 x 3 = 21 h
From relation,
E = P × t
Substituting the values in the formula above, we get,
E = 2 x 21 = 42 kWh
cost for 1 kWh of energy = ₹ 4.25
∴ Cost for 42 kWh of energy = 4.25 x 42
= ₹178.50
Answer:
(i) Given,
P = 1500 W
V = 250 volt
Current drawn (I) =
Substituting the values in the formula, we get,
I =
⇒ I = 6 A
Hence, current drawn = 6 A
(ii) energy consumed in 50 hours = ?
t = 50 h
From relation,
E = P × t
Substituting the values in the formula above, we get,
E = 1500 x 50 = 75000 Wh = 75 kWh
Hence, energy consumed in 50 hours = 75 kWh
(iii) cost of energy consumed at ₹ 4.20 per kWh = ?
Cost per unit of energy = ₹ 4.20
Cost for 75 kWh of energy = 4.2 x 7.5
= ₹ 315
Hence, cost of energy consumed = ₹ 315
Exercise 8C Short Questions
8 questionsAnswer:
(a) In the expression,
Symbol Q represents charge and the symbol V represents voltage.
(b) The expression for power P in terms of current and resistance is:
where,
I represents current and
R represents resistance.
Answer:
(i) The household unit of electricity is Kilowatt hour (kWh).
One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kilowatt when it is used for 1 hour.
(ii) The voltage of the electricity that is generally supplied to a house is 220 volt.
(iii) The electrical energy is consumed by various appliances in our houses (or industries) and it's cost is paid to the electrical company.
Answer:
One kilowatt-hour (kwh) is the electrical energy consumed by an electrical appliance of power 1 kilowatt when it is used for 1 hour.
Value of kilowatt-hour (kWh) in S.I. unit is —
1 kWh = 3.6 × 106 J
Answer:
Kilowatt is the unit of electrical power. i.e., one kilowatt is the electric power consumed when a current of 1 ampere flows through a circuit having a potential difference of 1 volt.
Kilowatt-hour is the unit of electrical energy i.e., One kilowatt - hour (kwh) is the electrical energy consumed by an electrical appliance of power 1 kilowatt when it is used for 1 hour.
Answer:
An electrical appliance such as electric heater, geyser etc is rated with it's power and voltage.
For example, an electric bulb is rated as 100 W - 220 V. It means that if the bulb is lighted on a 220 V supply, the electric power consumed by it is 100 W (i.e., 100 J of electric energy is consumed by the bulb in 1 s).
(a) Resistance of element of appliance while in use is —
(b) The safe limit of current in it, while in use is —
If current exceeds this value, the power supplied at voltage V will exceed the rated power of the appliance and the appliance may get damaged. So this value of current is called the safe current which can flow through the appliance at voltage V.
Answer:
When an electric bulb is rated '100 W, 250 V' it means that if the bulb is lighted on a 250 V supply, the electric power consumed by it is 100 W (i.e., 100 J of electrical energy is consumed by the bulb in 1 s).
Exercise 8C Very Short Questions
6 questionsAnswer:
The expression for electrical energy spent in flow of current through an electrical appliance in terms of current, resistance and time is:
W = I2Rt
Answer:
(a) Expression for electrical power spent in flow of current through a conductor in terms of resistance and potential difference is —
(b) Expression for electrical power spent in flow of current through a conductor in terms of current and resistance is —
Answer:
(i) The physical quantity measured in kW is electrical power.
(ii) The physical quantity measured in kWh is electrical energy. kWh is commercial unit of electrical energy.
(iii) The physical quantity measured in Wh is electrical energy. Wh is commercial unit of electrical energy.
Answer:
For circuit (I) :
Net Resistance (R) = 3 Ω
Voltage (V) = 6 V
Heat Dissipated = P = W .......... (1)
For circuit (II) :
As resistances are in parallel combination and also have equal value then
So, net resistance (R) = Ω
Voltage (V) = 6V
Heat Dissipated = P = W .......... (2)
For circuit (III) :
As resistances are in series combination then
So, net resistance (R) = 5 Ω
Voltage (V) = 6V
Heat Dissipated = P = W .......... (3)
From (1), (2) and (3) it is clearly visible that in circuit (iii) heat dissipation is minimum.