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(C) The circuit consists of two resistors of $64 \, \Omega$ and $32 \, \Omega$ connected in series across a potential difference of $60 \, V$.Using th

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$20$

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(C) The circuit consists of two resistors of $64 \, \Omega$ and $32 \, \Omega$ connected in series across a potential difference of $60 \, V$.Using the voltage divider rule,the potential difference across the $32 \, \Omega$ resistor is given by:$V_J - V_G = \left( \frac{R_2}{R_1 + R_2} ight) 	imes V_{total}$$V_J - V_G = \left( \frac{32}{64 + 32} ight) 	imes 60 \, V$$V_J - V_G = \left( \frac{32}{96} ight) 	imes 60 \, V$$V_J - V_G = \left( \frac{1}{3} ight) 	imes 60 \, V = 20 \, V$Since the potential at $G$ is $V_G = 0 \, V$,the potential at $J$ is $V_J = 20 \, V$.

Find the potential of $J$ with respect to $G$ in $V$.

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