Current Electricity — Question 6

For circuit (I) :

Net Resistance (R) = 3 Ω

Voltage (V) = 6 V

Heat Dissipated = P = $\dfrac{V^2}{R} = \dfrac{6\times6}{3} =\dfrac{36}{3} = 12$ W .......... (1)

For circuit (II) :

As resistances are in parallel combination and also have equal value then

$\dfrac{1}{R} = \dfrac{1}{3} + \dfrac{1}{3} = \dfrac{2}{3}$

So, net resistance (R) = $\dfrac{3}{2}$ Ω

Voltage (V) = 6V

Heat Dissipated = P = $\dfrac{V^2}{R} = \dfrac{6\times6}{\dfrac{3}{2}} =\dfrac{36\times2}{3} = 12\times2=24$ W .......... (2)

For circuit (III) :

As resistances are in series combination then

$R = 2 + 3 = 5\ Ω$

So, net resistance (R) = 5 Ω

Voltage (V) = 6V

Heat Dissipated = P = $\dfrac{V^2}{R} = \dfrac{6\times6}{5} =\dfrac{36}{5} = 7.2$ W .......... (3)

From (1), (2) and (3) it is clearly visible that in circuit (iii) heat dissipation is minimum.