Question 1

Question 1A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio RR′\dfrac{	ext{R}}{	ext{R}'}R′R​ is ...............125\dfrac{1}{25}251​15\dfrac{1}{5}51​525

Accepted Answer

25Reason — Given,Wire of resistance R is cut into five equal parts hence, resistance of each part is R5\dfrac{\text{R}}{5}5R​1R′=1R5+1R5+1R5+1R5+1R5=5R+5R+5R+5R+5R=25R⇒R′=R25\dfrac{1}{\text{R}'} = \dfrac{1}{\dfrac{\text{R}}{5}} + \dfrac{1}{\dfrac{\text{R}}{5}}+ \dfrac{1}{\dfrac{\text{R}}{5}} + \dfrac{1}{\dfrac{\text{R}}{5}} + \dfrac{1}{\dfrac{\text{R}}{5}} \[1em] = \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} \[1em] = \dfrac{25}{\text{R}} \[1em] \Rightarrow \text{R}' = \dfrac{\text{R}}{25}R′1​=5R​1​+5R​1​+5R​1​+5R​1​+5R​1​=R5​+R5​+R5​+R5​+R5​=R25​⇒R′=25R​Hence,RR′=RR25⇒RR′=25\dfrac{\text{R}}{\text{R}'} = \dfrac{\text{R}}{\dfrac{\text{R}}{25}} \[1em] \Rightarrow \dfrac{\text{R}}{\text{R}'} = 25R′R​=25R​R​⇒R′R​=25

Question 2

Question 2Which of the following terms does not represent electrical power in a circuit?I2RIR2VIV2R\dfrac{	ext{V}^2}{	ext{R}}RV2​

Accepted Answer

IR2Reason — Electrical power in a circuit:P = VI = I2R = V2R\dfrac{	ext{V}^2}{	ext{R}}RV2​

Question 3

Question 3An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be ...............100 W75 W50 W25 W

Accepted Answer

25 WReason —Given, V = 220V ; P = 100 W ; R = ?We know,⇒P=V2R⇒R=V2P\phantom{\Rightarrow} \text{P} = \dfrac{\text{V}^2}{\text{R}} \[1em] \Rightarrow \text{R} = \dfrac{\text{V}^2}{\text{P}}⇒P=RV2​⇒R=PV2​Substituting the values, we getR = 2202100\dfrac{\text{220}^2}{\text{100}}1002202​ = 484 ΩWhen voltage is reduced, resistance remains the same. Hence, the power consumed :P = 1102484\dfrac{110^2}{484}4841102​ = 25 WHence, the power consumed = 25 W.

Question 4

Question 4Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be1:22:11:44:1

Accepted Answer

1:4Reason — Let Rs and Rp be the equivalent resistance of the wires when connected in series and parallel respectively.Rs = R + R = 2R1Rp=1R+1R=2R\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{R}} + \dfrac{1}{\text{R}} = \dfrac{2}{\text{R}}Rp​1​=R1​+R1​=R2​Rp = R2\dfrac{\text{R}}{2}2R​Let Ps and Pp be the power consumed in series and parallel circuits, respectively.Power (P) = V2R\dfrac{\text{V}^2}{\text{R}}RV2​Ps = V22R\dfrac{\text{V}^2}{\text{2R}}2RV2​andPp = V2R2=2V2R\dfrac{\text{V}^2}{\dfrac{\text{R}}{2}} = \dfrac{2\text{V}^2}{\text{R}}2R​V2​=R2V2​For the same potential difference V, the ratio of the heat produced in the circuit is given by ratio of Ps and Pp henceHs:Hp=V22R2V2R=V2×R2R×2V2=14\text{H}_s : \text{H}_p = \dfrac{\dfrac{\text{V}^2}{\text{2R}}}{\dfrac{2\text{V}^2}{\text{R}}} \[1em] = \dfrac{\text{V}^2 \times \text{R}}{\text{2R} \times 2\text{V}^2} \[1em] = \dfrac{1}{4}Hs​:Hp​=R2V2​2RV2​​=2R×2V2V2×R​=41​Hence, the ratio of the heat produced is 1:4.

Question 5

Question 5How is a voltmeter connected in the circuit to measure the potential difference between two points?

Accepted Answer

In the circuit, voltmeter is connected in parallel to measure the potential difference between two points.

Question 6

Question 6A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Accepted Answer

Given,diameter = 0.5 mmarea = πd24=3.14×(5×10−4)24\dfrac{πd^2}{4} = \dfrac{3.14 	imes (5 	imes 10^{-4})^2}{4}4πd2​=43.14×(5×10−4)2​ = 19.625 x 10–8 mresistivity = 1.6 × 10–8 Ω mR = 10 Ωl = ?We know, R = ρ lA\dfrac{l}{A}Al​Substituting we get,10 = 1.6 × 10–8 x l19.625×10−8\dfrac{l}{19.625 	imes 10^{-8}}19.625×10−8l​l = 10×19.6251.6\dfrac{10 	imes 19.625}{1.6 }1.610×19.625​ = 122.7 mThe length of the wire is 122.72 m.When the diameter is doubledAccording to the formula,R = ρ lA\dfrac{l}{A}Al​ where A = πd24\dfrac{πd^2}{4}4πd2​Hence, R ∝ 1d2\dfrac{1}{d^2}d21​When diameter is double,Resistance will change by 1(2d)2\dfrac{1}{(2d)^2}(2d)21​ = 14d2\dfrac{1}{4d^2}4d21​∴ Resistance will become 14th\dfrac{1}{4}^{	ext{th}}41​th of its original value.Hence, new resistance = 14×10\dfrac{1}{4} 	imes 1041​×10 = 2.5 Ω

Question 7

Question 7The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –I (ampere)0.51.02.03.04.0V (volts)1.63.46.710.213.2Plot a graph between V and I and calculate the resistance of that resistor.

Accepted Answer

The graph is plotted as below:The slope of the line indicates the value of resistance.Let P1(3.4, 1) and P2(10.2, 3) be two points on the graphSlope = 1R=3−110.2−3.4=26.8\dfrac{1}{	ext{R}} = \dfrac{3-1}{10.2-3.4} = \dfrac{2}{6.8}R1​=10.2−3.43−1​=6.82​R = 6.82\dfrac{6.8}{2}26.8​ = 3.4 ΩTherefore, the resistance = 3.4 Ω.

Question 8

Question 8When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Accepted Answer

Given,V = 12 VI = 2.5 mA = 2.5 x 10-3AR = ?According to Ohm's Law:V = IR∴R=VI⇒R=122.5×10−3⇒R=4.8×103Ω\therefore \text{R} = \dfrac{\text{V}}{\text{I}} \[1em] \Rightarrow \text{R} = \dfrac{12}{2.5 \times 10^{-3}} \[1em] \Rightarrow \text{R} = 4.8 \times 10^{3} \Omega∴R=IV​⇒R=2.5×10−312​⇒R=4.8×103ΩHence, resistance of the resistor = 4.8 kΩ

Question 9

Question 9A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Accepted Answer

Given,V = 9VResistors in series = 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω,Rs = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 ΩAccording to Ohm's Law:V = IR∴I=VR⇒I=913.4⇒I=0.67 A\therefore \text{I} = \dfrac{\text{V}}{\text{R}} \[1em] \Rightarrow \text{I} = \dfrac{9}{13.4} \[1em] \Rightarrow \text{I} = 0.67 \text{ A}∴I=RV​⇒I=13.49​⇒I=0.67 ATherefore, current flowing = 0.67 A.

Question 10

Question 10How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Accepted Answer

Given,I = 5 AV = 220 VLet number of resistors be nResistance of each resistor = 176 Ω1Req=1176+1176+....n1Req=n176Ω⇒Req=176nΩ\dfrac{1}{\text{R}_{eq}} = \dfrac{1}{176} + \dfrac{1}{176} + ....n \[1em] \dfrac{1}{\text{R}_{eq}} = \dfrac{\text{n}}{176} \Omega \[1em] \Rightarrow \text{R}_{eq} = \dfrac{176}{\text{n}} \OmegaReq​1​=1761​+1761​+....nReq​1​=176n​Ω⇒Req​=n176​ΩApplying Ohm's Law:V = IR220 = 5 x 176n\dfrac{176}{\text{n}}n176​n = 176×5220\dfrac{176 \times 5}{220}220176×5​ = 4 resistorsHence, 4 resistors are connected in parallel.

Question 11

Question 11Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Accepted Answer

If we connect resistors in series, R = 6 Ω + 6 Ω + 6 Ω = 18 Ω.If we connect all resistors in parallel,1R=16+16+16=36=12\dfrac{1}{	ext{R}} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{3}{6} = \dfrac{1}{2}R1​=61​+61​+61​=63​=21​Hence, R = 2ΩWe can obtain the desired value by connecting two of the resistors in either series or parallel.Case (i)If two resistors are connected in parallel, then their equivalent resistance is1Rp=16+16=26=13\dfrac{1}{	ext{R}_	ext{p}} = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}Rp​1​=61​+61​=62​=31​Hence, Rp = 3ΩThe third resistor in series, then we get equivalent resistanceRp + 6 = 3 + 6 = 9Ω∴ To get a resistance of 9 Ω, two 6 Ω resistors should be connected in parallel and the third 6 Ω resistor should be connected in series with the combination as shown below:Case (ii)When two resistors are connected in series, their equivalent resistance isRs = 6 Ω + 6 Ω = 12 ΩThe third resistor is connected in parallel with 12 Ω. Hence the equivalent resistance is:1R=112+16=1+212=312=14\dfrac{1}{	ext{R}} = \dfrac{1}{12} + \dfrac{1}{6} = \dfrac{1+2}{12} = \dfrac{3}{12} = \dfrac{1}{4}R1​=121​+61​=121+2​=123​=41​⇒ R = 4 Ω∴ To get a resistance of 4 Ω, two 6 Ω resistors should be connected in series and the third 6 Ω resistor should be connected in parallel with the combination as shown below:

Question 12

Question 12Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Accepted Answer

Given,V = 220 VP = 10 WIMax = 5 AThe resistance of the each bulb:R = V2P=220210=4840Ω\dfrac{\text{V}^2}{\text{P}} = \dfrac{220^2}{10} = 4840 ΩPV2​=102202​=4840ΩAssuming n bulbs are connected in parallel1Req=1R+1R+....n1Req=14840+14840+....n1Req=n4840Ω⇒Req=4840nΩ\dfrac{1}{\text{R}_{eq}} = \dfrac{1}{\text{R}} + \dfrac{1}{\text{R}} + ....n \[1em] \dfrac{1}{\text{R}_{eq}} = \dfrac{1}{4840} + \dfrac{1}{4840} + ....n \[1em] \dfrac{1}{\text{R}_{eq}} = \dfrac{\text{n}}{4840} \Omega \[1em] \Rightarrow \text{R}_{eq} = \dfrac{4840}{\text{n}} \OmegaReq​1​=R1​+R1​+....nReq​1​=48401​+48401​+....nReq​1​=4840n​Ω⇒Req​=n4840​ΩApplying Ohm's Law:V = IR220 = 5 x 4840n\dfrac{4840}{\text{n}}n4840​n = 4840×5220\dfrac{4840 \times 5}{220}2204840×5​ = 110 lampsHence, 110 lamps can be connected in parallel.

Question 13

Question 13A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Accepted Answer

Case 1 — When coils are used separatelyApplying Ohm's law:V=IR∴I=VR⇒I=22024⇒I=9.166 A≈9.2 A\text{V} = \text{IR} \[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \[1em] \Rightarrow \text{I} = \dfrac{220}{24} \[1em] \Rightarrow \text{I} = 9.166 \text{ A} \approx 9.2 \text{ A}V=IR∴I=RV​⇒I=24220​⇒I=9.166 A≈9.2 AHence, 9.2 A of current flows through each resistor when they are used separately.Case 2 — When coils are connected in seriesRs = 24 Ω + 24 Ω = 48 ΩApplying Ohm's law:V=IR∴I=VR⇒I=22048⇒I=4.58 A≈4.6 A\text{V} = \text{IR} \[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \[1em] \Rightarrow \text{I} = \dfrac{220}{48} \[1em] \Rightarrow \text{I} = 4.58 \text{ A} \approx 4.6 \text{ A}V=IR∴I=RV​⇒I=48220​⇒I=4.58 A≈4.6 ATherefore, the current is 4.6 A when the two coils are used in series.Case 3 — When coils are connected in parallel1Rp=124+124=224=112\dfrac{1}{\text{R}_p} = \dfrac{1}{24} + \dfrac{1}{24} = \dfrac{2}{24} = \dfrac{1}{12}Rp​1​=241​+241​=242​=121​∴ Rp = 12 ΩApplying Ohm's law:V=IR∴I=VR⇒I=22012⇒I=18.3 A\text{V} = \text{IR} \[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \[1em] \Rightarrow \text{I} = \dfrac{220}{12} \[1em] \Rightarrow \text{I} = 18.3 \text{ A}V=IR∴I=RV​⇒I=12220​⇒I=18.3 ATherefore, the current is 18.3 A when the two coils are used in parallel.

Question 14

Question 14Compare the power used in the 2 Ω resistor in each of the following circuits:(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Accepted Answer

(i) The potential difference is 6 V and the resistors 1 Ω and 2 Ω are connected in series:Rs = 1 Ω + 2 Ω = 3 Ω.Applying Ohm's law:V=IR∴I=VR⇒I=63⇒I=2 A\text{V} = \text{IR} \[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \[1em] \Rightarrow \text{I} = \dfrac{6}{3} \[1em] \Rightarrow \text{I} = 2 \text{ A}V=IR∴I=RV​⇒I=36​⇒I=2 APower across 2 Ω resistor = I2R = 2 x 2 x 2 = 8 WHence, the power consumed by the 2 Ω resistor is 8 W.(ii) In parallel circuit, voltage across the resistors remains the same. So, voltage across 2 Ω resistor is 4 V:Power across 2 Ω resistor = V2R=4×42\dfrac{\text{V}^2}{{\text{R}}} = \dfrac{4 \times 4}{2}RV2​=24×4​ = 8 WThe power consumed by the 2 Ω resistor is 8 W.Hence, same power is consumed by 2 Ω resistor in both the circuits.

Question 15

Question 15Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Accepted Answer

As the two lamps are connected in parallel, the voltage across each of them will be the same.Let I1 be the current drawn by 100 W bulb:I1 = PV=100220=511 A\dfrac{\text{P}}{\text{V}} = \dfrac{100}{220} = \dfrac{5}{11} \text{ A}VP​=220100​=115​ ALet I2 be the current drawn by 60 W bulb:I2 = PV=60220=311 A\dfrac{\text{P}}{\text{V}} = \dfrac{60}{220} = \dfrac{3}{11} \text{ A}VP​=22060​=113​ A∴ Current drawn from line = I1 + I2=511+311=5+311=811 A=0.727272 A≈0.73 A= \dfrac{5}{11} + \dfrac{3}{11} \[1em] = \dfrac{5 + 3}{11} \[1em] = \dfrac{8}{11} \text{ A} \[1em] = 0.727272 \text{ A} \approx 0.73 \text{ A}=115​+113​=115+3​=118​ A=0.727272 A≈0.73 AHence, 0.73 A current flows through the line.

Question 16

Question 16Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Accepted Answer

Given,Power of TV set = P1 = 250 WTime taken by TV set = t1 = 1 hr = 3600 sPower of toaster = P2 = 1200Time taken by toaster = t2 = 10 minutes = 600 sWe know, energy consumed (H) = PtEnergy consumed by TV set:E1 = 250 × 3600 = 900000 J = 9 × 105 JEnergy consumed by toaster:E2 = 1200 × 600 = 720000 = 7.2 × 105 JAs E1 > E2, hence, energy consumed by the TV set is greater than the toaster.

Question 17

Question 17An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Accepted Answer

Given,I = 15 AR = 8 ΩThe rate at which the heat develops in the heater (P) = I2RSubstituting the values in the equation, we getP = 15 x 15 x 8 = 1800 wattTherefore, electric heater produces heat at the rate of 1800 W

Question 18

Question 18Explain the following.(a) Why is the tungsten used almost exclusively for filament of electric lamps?(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?(c) Why is the series arrangement not used for domestic circuits?(d) How does the resistance of a wire vary with its area of cross-section?(e) Why are copper and aluminium wires usually employed for electricity transmission?

Accepted Answer

(a) The resistivity and melting point of tungsten is very high hence it doesn't burn readily when heated. Therefore, tungsten is used almost exclusively for filament of electric lamps.(b) Resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidize (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices like electric iron, toasters etc.(c) The series arrangement is not used for domestic circuits due to the following reasons:All the connected appliances cannot be operated independently. If one device is defective, then the entire circuit will not function.The overall voltage gets distributed in a series circuit. As a result, electric appliances may not get the rated power for their operation.The total resistance becomes large, hence, the current is reduced.(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A) i.e., when the area of cross section increases the resistance decreases and vice versa.R = ρ lA\dfrac{l}{A}Al​(e) Copper and aluminium both are good conductors of electricity and have low resistivity. Hence, they can efficiently carry electrical current with minimal resistance, which reduces energy losses as heat during transmission.

Question 19

Question 1What does an electric circuit mean?

Accepted Answer

A continuous and closed path of an electric current is called an electric circuit.

Question 20

Question 2Define the unit of current.

Accepted Answer

The unit of current is ampere (A). One ampere is constituted by the flow of one coulomb of charge per second, i.e., 1A = 1C1s\dfrac{1 	ext{C}}{1 	ext{s}}1s1C​

Electricity

Exercises

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