Q: Question 2An electric iron of power 1.5 kW is used for 30 minutes to press clothes. Calculate the electrical energy consumed in:(a) kilowatt hour(b) joule

Power (P) = 1.5 kWtime (t) = 30 minutes = 12\dfrac{\text{1}}{\text{2}}21 hr(a) Electrical energy consumed in kWh= P x t= 1.5 x 12\dfrac{\text{1}}{\text{2}}21= 0.75 kWhSo electrical energy consumed in kWh = 0.75 kWh(b) 1 kWh = 3.6 x 106 J0.75 kWh = 0.75 x 3.6 x 106= 2.7 x 106 JSo electrical energy consumed in joule = 2.7 x 106 J.

Question 1

Question 1An electrical appliance is rated as 60 W - 150 V.(a) What do you understand by this statement?(b) How much current will flow through the appliance when in use?

Accepted Answer

(a) If an electrical appliance is rated as 60 W - 150 V, it means if the appliance is put on a 150 V supply, it will consume 60 W electrical power (i.e., it will consume 60 J of electrical energy in 1 s).(b) P = 60 WV = 150 VWe know P = VI∴ I = PV\dfrac{	ext{P}}{	ext{V}}VP​ = 60150\dfrac{	ext{60}}{	ext{150}}15060​ = 0.4 ASo, current of 0.4 A will flow through the appliance when in use.

Question 2

Question 2An electric iron of power 1.5 kW is used for 30 minutes to press clothes. Calculate the electrical energy consumed in:(a) kilowatt hour(b) joule

Accepted Answer

Power (P) = 1.5 kWtime (t) = 30 minutes = 12\dfrac{	ext{1}}{	ext{2}}21​ hr(a) Electrical energy consumed in kWh= P x t= 1.5 x 12\dfrac{	ext{1}}{	ext{2}}21​= 0.75 kWhSo electrical energy consumed in kWh = 0.75 kWh(b) 1 kWh = 3.6 x 106 J0.75 kWh = 0.75 x 3.6 x 106= 2.7 x 106 JSo electrical energy consumed in joule = 2.7 x 106 J.

Question 3

Question 3Assuming the electric consumption per day to be 12 kWh and the rate of electricity to be Rs 6.25 per unit, find how much money is to be paid in a month of 30 days?

Accepted Answer

Electric energy consumed per day = 12 kWhElectric energy consumed for 30 days = 12 x 30 = 360 kWh1 kWh = 1 unit360 kWh = 360 units1 unit cost = Rs 6.25360 units cost = 360 x 6.25 = Rs 2250So money to be paid for 30 days = Rs 2250.

Question 4

Question 4In a premise 5 bulbs each of 100 W, 2 fans each of 60 W, 2 A.Cs each of 1.5 kW are used for 5 h per day. Find:(a) total power consumed per day,(b) total power consumed in 30 days,(c) total electrical energy consumed in 30 days,(d) the cost of electricity at the rate of Rs 6.25 per unit.

Accepted Answer

(a) Power consumed by 5 bulbs = 5 x 100 = 500 WPower consumed by 2 fans = 2 x 60 = 120 WPower consumed by 2 A.Cs = 2 x 1.5 = 3 kW or 3 x 1000 = 3000 WTotal power consumed per day = 500 + 120 + 3000 = 3620 WTotal power consumed per day = 3620 W.(b) Power consumed for 30 days = 30 x 3620 = 108600 W or 108.6 kWPower consumed for 30 days = 108.6 kW.(c) Power consumed in 30 days = 108.6 kWtime = 5 hElectrical energy consumed in 30 days = Power x time= 108.6 x 5= 543 kWhElectrical energy consumed in 30 days = 543 kWh.(d) Cost of 1 unit = Rs 6.25Cost of 543 units = 6.25 x 543 = Rs 3393.75Cost of electricity = Rs 3393.75.

Question 5

Question 6One day the meter reading was 7643 units while on the following day, it was 7657 units. What is the consumption of electricity in a day?

Accepted Answer

Initial reading = 7643 unitsFinal reading = 7657 unitsConsumption = Final reading - Initial reading= 7657 - 7643= 14 unitsSo, consumption of electricity in one day is 14 units.

Question 6

Question 7A source of potential difference V volt sends current I ampere in a circuit for time t second. Write expressions for(a) electrical energy supplied by the source, and(b) electrical power spent by the source.

Accepted Answer

(a) Potential difference = V voltCurrent = I amperetime = t secElectrical energy supplied W = VIt(b) Power is the rate of doing work.P = Wt\dfrac{	ext{W}}{	ext{t}}tW​ = VItt\dfrac{	ext{VIt}}{	ext{t}}tVIt​ = VIP = VI

Question 7

Question 8Name the unit in which you pay the cost of your electricity bill. How is it related to joule?

Accepted Answer

We pay the cost of our electricity bill in unit kWh (kilowatt hour).Since,Power=EnergyTime∴Energy=Power×Timei.e. Energy (in kWh)=Power (in kWh)×Time (in h)\text{Power} = \dfrac{\text{Energy}}{\text{Time}} \[1em] \therefore \text{Energy} = \text{Power} \times \text{Time} \[1em] \text{i.e. } \text{Energy (in kWh)} \[1em] = \text{Power (in kWh)} \times \text{Time (in h)}Power=TimeEnergy​∴Energy=Power×Timei.e. Energy (in kWh)=Power (in kWh)×Time (in h)Hence,1 kWh = 1 kW x 1 h= 1000 W x (60 x 60 s)= 36,00,000 J= 3.6 x 106 JSo 1 kWh = 3.6 x 106 J.

Question 8

Question 9If an appliance of power P watt is used for time t hour, how much electrical energy is consumed in kWh?

Accepted Answer

Power = P watt or P x 10-3 kWtime = t hourPower = EnergyTime\dfrac{	ext{Energy}}{	ext{Time}}TimeEnergy​Energy = Power x time= P x 10-3 kW x t h= P x t x 10-3 kWhElectrical energy consumed = Pt x 10-3 kWh.

Question 9

Question 10State one property of the material of a fuse wire.

Accepted Answer

Fuse wire is a short wire with a low melting point. It melts at about 200°C.

Question 10

Question 11Name the material of a fuse wire.

Accepted Answer

Fuse wire is an alloy of lead and tin, having 50% of each metal.

Question 11

Question 12How does a fuse protect the electric wiring (or an appliance) from being damaged?

Accepted Answer

During short circuiting or voltage fluctuations maximum current passes through the wires causing burning of the wires. To prevent this damage a fuse is connected to live wire of the circuit. When there is a short circuit, the fuse wire gets heated up to the extent that it melts. As a result, a gap is produced in the live wire and the circuit breaks. No current then flows and electrical appliance or circuit is protected from being damaged.

Question 12

Question 13Write the full form of M.C.B.

Accepted Answer

The full form of M.C.B. is miniature circuit breaker.

Question 13

Question 14How is M.C.B. superior to the fuse wire?

Accepted Answer

M.C.B. is superior to the fuse wire in the following ways:An M.C.B. switches off the circuit in a very short time (nearly 25 ms) as compared to a fuse wire which takes relatively longer to melt and break the circuit.M.C.B. can be reset (i.e., raised up) after tripping, whereas fuse wire needs to be replaced after it has been tripped.M.C.B. offers greater protection, flexibility, and safety compared to fuse wire.

Question 14

Question 15With which wire: live or neutral is the fuse wire connected?

Accepted Answer

The fuse wire is connected to the live wire.

Question 15

Question 16State the voltage at which electricity is supplied to our houses.

Accepted Answer

Electricity is supplied to our houses at 220 volt.

Question 16

Question 17State the colour coding of the three wires in a cable used for wiring in a household electrical circuit.

Accepted Answer

In the cable there are three wires live, neutral and earth wire. Live wire is red or brown. Neutral wire is black or light blue. Earth wire is green or yellow.

Question 17

Question 18Why is the metal covering of an electrical appliance earthed?

Accepted Answer

The metal covering of an electrical appliance is connected to the earth wire to protect from electric shock. Sometimes due to breaking of insulation of wires, live wires come in contact with the body of the appliance which can cause a fatal shock when touched. If the appliance is earthed, the current will pass to the earth and the person remains protected from getting any electrical shock.

Question 18

Question 5How are the electrical appliances connected in a house circuit: in series or in parallel? Give reason.

Accepted Answer

The electrical appliances are connected in parallel in a house circuit due to following reasons :Each appliance works independently. If one appliance in the circuit stops working, it doesn't affect the working of other appliances.Each appliance has an independent path for current and works at the same voltage.Each appliance can have a separate switch connected to its live wire.

Household Electricity

Exercise 8A Numericals

Exercise 8A Short Answer Type Questions

Exercise 8A Think And Answer