Three identical resistors R_1 = R_2 = R_3 = R | vedclass

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(B) Let the resistance of each resistor be $R$. In the given circuit,$R_2$ and $R_3$ are connected in parallel. Their equivalent resistance is $R_p =

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Greatest in $R_1$

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(B) Let the resistance of each resistor be $R$. In the given circuit,$R_2$ and $R_3$ are connected in parallel. Their equivalent resistance is $R_p = \frac{R 	imes R}{R + R} = \frac{R}{2}$. This parallel combination is in series with $R_1$. The total resistance of the circuit is $R_{eq} = R_1 + R_p = R + \frac{R}{2} = \frac{3R}{2}$. The total current flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{V}{3R/2} = \frac{2V}{3R}$. This entire current $I$ flows through $R_1$. Thus,the power dissipated in $R_1$ is $P_1 = I^2 R = (\frac{2V}{3R})^2 R = \frac{4V^2}{9R}$. The current $I$ splits equally between $R_2$ and $R_3$ because they are identical. So,the current through $R_2$ (and $R_3$) is $I_2 = I_3 = \frac{I}{2} = \frac{V}{3R}$. The power dissipated in $R_2$ is $P_2 = I_2^2 R = (\frac{V}{3R})^2 R = \frac{V^2}{9R}$. Comparing the powers,$P_1 = \frac{4V^2}{9R}$ and $P_2 = P_3 = \frac{V^2}{9R}$. Clearly,$P_1 > P_2 = P_3$. Therefore,the power dissipated is greatest in $R_1$.

Three identical resistors $R_1 = R_2 = R_3 = R$ are connected as shown to a battery of constant e.m.f. $V$. The power dissipated is:

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