The three resistances A, B and C have values 3R, 6R and R respectively. When some potential difference is applied across the network, the thermal powers dissipated by A, B and C are in the ratio

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2 : 3 : 4

2 : 4 : 3

4 : 2 : 3

3 : 2 : 4

answer is C.

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Detailed Solution

Thermal power in

$A = P_{A} = {(\frac{2 i}{3})}^{2} 3 R = \frac{4}{3} i^{2} R$

Thermal power in

$B = P_{B} = {(\frac{i}{3})}^{2} 6 R = \frac{2}{3} i^{2} R$

Question Image

Thermal power in $C = P_{C} = i^{2} R$

$\Rightarrow P_{A} : P_{B} : P_{C}$ $= \frac{4}{3} : \frac{2}{3} : 1 = 4 : 2 : 3$

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