The three resistances A, B and C have values | vedclass

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(C) Let the total current flowing through the series combination of the parallel network and resistor $C$ be $i$. The current $i$ flows through resist

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$4 : 2 : 3$

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(C) Let the total current flowing through the series combination of the parallel network and resistor $C$ be $i$. The current $i$ flows through resistor $C$. Resistors $A$ $(3R)$ and $B$ $(6R)$ are in parallel. Let $i_A$ and $i_B$ be the currents through $A$ and $B$ respectively. Using the current divider rule: $i_A = i 	imes \frac{6R}{3R + 6R} = i 	imes \frac{6R}{9R} = \frac{2}{3}i$ $i_B = i 	imes \frac{3R}{3R + 6R} = i 	imes \frac{3R}{9R} = \frac{1}{3}i$ Thermal power dissipated is given by $P = I^2 R$. Power in $A$: $P_A = (i_A)^2 (3R) = (\frac{2}{3}i)^2 (3R) = \frac{4}{9} i^2 (3R) = \frac{4}{3} i^2 R$ Power in $B$: $P_B = (i_B)^2 (6R) = (\frac{1}{3}i)^2 (6R) = \frac{1}{9} i^2 (6R) = \frac{2}{3} i^2 R$ Power in $C$: $P_C = i^2 R$ The ratio $P_A : P_B : P_C = \frac{4}{3} i^2 R : \frac{2}{3} i^2 R : i^2 R = \frac{4}{3} : \frac{2}{3} : 1 = 4 : 2 : 3$.

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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