The resistances in the following circuit are | vedclass

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(A) Case $1$: When switch $S$ is open, $R_1$ and $R_3$ are in series. The total resistance is $R_{eq} = R_1 + R_3 = 6 + 6 = 12 \ \Omega$.The total cur

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$-2 \ V$

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(A) Case $1$: When switch $S$ is open, $R_1$ and $R_3$ are in series. The total resistance is $R_{eq} = R_1 + R_3 = 6 + 6 = 12 \ \Omega$.The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{12}{12} = 1 \ A$.The potential difference across $R_1$ is $V_1 = I 	imes R_1 = 1 	imes 6 = 6 \ V$.Case $2$: When switch $S$ is closed, $R_1$ and $R_2$ are in parallel, and this combination is in series with $R_3$.The equivalent resistance of the parallel part is $R_p = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{6 	imes 6}{6 + 6} = 3 \ \Omega$.The total resistance of the circuit is $R_{eq}' = R_3 + R_p = 6 + 3 = 9 \ \Omega$.The total current from the battery is $I' = \frac{V}{R_{eq}'} = \frac{12}{9} = \frac{4}{3} \ A$.The potential difference across the parallel combination (which is the potential across $R_1$) is $V_2 = I' 	imes R_p = \frac{4}{3} 	imes 3 = 4 \ V$.The change in potential across $R_1$ is $\Delta V = V_2 - V_1 = 4 - 6 = -2 \ V$.

The resistances in the following circuit are $R_1 = R_2 = R_3 = 6.0 \ \Omega$. The emf of the battery is $12 \ V$. When switch $S$ is closed, the potential across resistance $R_1$ changes by an amount:

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