The document Solutions of Electricity (Page No- 49 & 58) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Physics Solutions By Lakhmir Singh & Manjit Kaur.

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**Lakhmir Singh Physics Class 10 Solutions ****Page No:49**

**Question 18:****The figure below shows a variable resistor in a dimmer switch.**

**How would you turn the switch to make the lights : (a) brighter, and (b) dimmer ? Explain your answer.****Solution :**

(a) Turn the switch to right side so as the resistance decreases.

(b) Turn the switch to the left side so as the resistace increases.

**Lakhmir Singh Physics Class 10 Solutions ****Page No:58**

**Question 1:****State two factors on which the electrical energy consumed by an electrical appliance depends.****Solution :**

Electrical energy consumed by an electrical appliance depends on:

1. Power rating of the appliance.

2. Time for which the appliance is used.

**Question 2:****Which one has a higher electrical resistance : a 100 watt bulb or a 60 watt bulb ?****Solution :**

60 watt bulb, because power is inversely proportional to the resistance.

**Question 3:****Name the commercial unit of electric energy.****Solution :**

Kilowatt-hour is the commercial unit of electric energy.

**Question 4:****An electric bulb is rated at 220 V, 100 W. What is its resistance ?****Solution :**

V = 220 V, P = 100W

R=?

We know that

P = V2/R

Thus

R = V2/P = 2202/100 = 484ohm

**Question 5:****What is the SI unit of (i) electric energy, and (ii) electric power ?****Solution :**

(i) joule

(ii) watt

**Question 6:****Name the quantity whose unit is (i) kilowatt, and (ii) kilowatt-hour.****Solution :**

(i) Electric power

(ii) Electric energy

**Question 7:****Which quantity has the unit of watt ?****Solution :**

Electric power has the unit of watt.

**Question 8:****What is the meaning of the symbol kWh ? Which quantity does it represent ?****Solution :**

kWh is the short form of kilowatt-hour, which is the commercial unit of electrical energy.

**Question 9:****If the potential difference between the end of a wire of fixed resistance is doubled, by how much does the electric power increase ?****Solution :**

P = V2/R

R is fixed.

V becomes double.

Now, P = (2V)2/R = 4 V2/R

So, the electric power becomes four times its previous value.

**Question 10:****An electric lamp is labelled 12 V, 36 W. This indicates that it should be used with a 12 V supply. What other information does the label provide ? Solution :**

Other information is that it will consume energy at the rate of 36 J/s.

**Question 11:****What current will be taken by a 920 W appliance if the supply voltage is 230 V ?****Solution :**

P = 920W, V = 230V, I = ?

We know that

P = V x I,

920 = 230 x I

I = 920/230 = 4amp

**Question 12:****Define watt. Write down an equation linking watts, volts and amperes.****Solution :**

When an electrical appliance consumes electrical energy at the rate of 1 joule per second, its power is said to be 1 watt.

1 watt = 1 volt x 1 ampere.

**Question 13:****Define watt-hour. How many joules are equal to 1 watt-hour ?****Solution :**

One watt hour is the amount of electrical energy consumed when an electrical appliance of 1 watt power is used for 1 hour.

1 watt hour = 3600 joules

**Question 14:****How much energy is consumed when a current of 5 amperes flows through the filament (or element) of a heater having resistance of 100 ohms for two hours ? Express it in joules.****Solution :**

I = 5amp, R = 100ohms, t = 2h

We know that

Electric energy consumed = P x t = I*I*Rt

= 25 x 100 x 2

= 5000 Wh

= 5 kwh

We know that 1kwh = 3.6 x 106 J

Therefore, 5kwh = 5 x 3.6 x 106 J = 18 x 106 J.

**Question 15:****An electric bulb is connected to a 220 V power supply line. If the bulb draws a current of 0.5 A, calculate the power of the bulb.****Solution :**

V = 220V, I = 0.5amp, P = ?

We know that

P = VI = 220X0.5

P =110 watt.

**Question 16:****In which of the following cases more electrical energy is consumed per hour ? (i) A current of 1 ampere passed through a resistance of 300 ohms. (ii) A current of 2 amperes passed through a resistance of 100 ohms.**

(i) R = 300 ohm. I = 1 A. t = 1h

P = I^{2}R = 1^{2} x 300 = 300 W

E= Pxt= 300 x 1 = 300 Wh

(ii) R = 100 ohm. I = 2 A. t = 1h

P = I^{2}R = 2^{2} x 100 = 400 W

E = Pxt = 400 x 1 = 400 Wh

Hence. in case (ii). the electrical energy consumed per hour is more.

**Question 17:****An electric kettle rated at 220 V, 2.2 kW, works for 3 hours. Find the energy consumed and the current drawn.****Solution :**

V = 220V, P = 2.2kW = 2200W, t = 3h

We know that

Electrical energy consumed = Pxt = 2.2Ã—3 = 6.6 kWh

We have, P = V x I

2200 = 220 x I

I = 10amp

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