An 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.

Given,

Resistance of electric heater, R = 8 Ω

Current drawn by the heater, I= 15 A
Time, t = 2 h

The rate at which heat is developed in the heater is equivalent to the power.

∴ Power, P = I²R
                 = (15)² x 8
                 = 1800 Js^–1. 

Voltage, V = 220 V
Resistance of the coil A and B = 24 Ω 

When the two coils A and B are used separately, 
R = 24 Ω, V = 220 V 

Current, I = $\frac{\mathrm{V}}{\mathrm{R}} = \frac{220 \mathrm{V}}{24 \mathrm{\Omega }} = 9.167 \mathrm{A}.$

(ii) When the two coils are connected in series,

R = 24 + 24 = 48 Ω, V = 220 V

Current,  I = $\frac{\mathrm{V}}{\mathrm{R}} = \frac{220}{48} = 4.58 \mathrm{A}.$ 

(iii) When the two coils are connected in parallel,

$\mathrm{R} = \frac{24 \times 24}{24+24} = 12 \mathrm{\Omega }, \mathrm{V} = 220 \mathrm{V}$ 

Current, I = $\frac{\mathrm{V}}{\mathrm{R}} = \frac{220}{12} = 18.33 \mathrm{A}.$ 

(i) The circuit diagram is shown in Fig.

Total resistance,   R = 1 + 2 = 3 Ω ,  Potential difference, V = 6 V
Current,   I = 
Power used in 2 Ω resistor = I₂R = (2)² x 2 = 8.W.    Ans.
(ii) The circuit diagram for this case is shown below:

Power used in  resistor =