Three resistances A, B and C having values 3R | vedclass

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(C) Let the total potential difference applied across the network be $V$. The resistors $A$ $(3R)$ and $B$ $(6R)$ are in parallel,and their equivalent

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$4: 2: 3$

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(C) Let the total potential difference applied across the network be $V$. The resistors $A$ $(3R)$ and $B$ $(6R)$ are in parallel,and their equivalent resistance $R_{AB}$ is given by $\frac{1}{R_{AB}} = \frac{1}{3R} + \frac{1}{6R} = \frac{2+1}{6R} = \frac{3}{6R} = \frac{1}{2R}$,so $R_{AB} = 2R$.This combination is in series with resistor $C$ $(R)$. The total resistance of the circuit is $R_{total} = R_{AB} + R_C = 2R + R = 3R$.The current flowing through the circuit is $I = \frac{V}{3R}$.The potential difference across the parallel combination ($A$ and $B$) is $V_{AB} = I 	imes R_{AB} = \frac{V}{3R} 	imes 2R = \frac{2V}{3}$.The potential difference across resistor $C$ is $V_C = I 	imes R_C = \frac{V}{3R} 	imes R = \frac{V}{3}$.The power dissipated in a resistor is given by $P = \frac{V^2}{R}$.For resistor $A$: $P_A = \frac{(2V/3)^2}{3R} = \frac{4V^2/9}{3R} = \frac{4V^2}{27R}$.For resistor $B$: $P_B = \frac{(2V/3)^2}{6R} = \frac{4V^2/9}{6R} = \frac{4V^2}{54R} = \frac{2V^2}{27R}$.For resistor $C$: $P_C = \frac{(V/3)^2}{R} = \frac{V^2/9}{R} = \frac{V^2}{9R} = \frac{3V^2}{27R}$.The ratio of powers $P_A : P_B : P_C$ is $\frac{4V^2}{27R} : \frac{2V^2}{27R} : \frac{3V^2}{27R} = 4 : 2 : 3$.

Three resistances $A, B$ and $C$ having values $3R, 6R$ and $R$ respectively form a network as shown. When a potential difference is applied across the network,the thermal powers dissipated by $A, B$ and $C$ are in the ratio:

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