Solutions of Electricity (Page No- 49 & 58) - Physics By Lakhmir Singh, Class 10 | Extra Documents, Videos & Tests for Class 10 PDF Download

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.
Solution :

(i) R = 300 ohm. I = 1 A. t = 1h
P = I²R = 1² x 300 = 300 W
E= Pxt= 300 x 1 = 300 Wh 

(ii) R = 100 ohm. I = 2 A. t = 1h
P = I²R = 2² 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 = 2.2×3 = 6.6 kWh
We have, P = V x I
2200 = 220 x I
I = 10amp