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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 at point $J$ with respect to ground $(G)$ is given by:$V_J = V_{total} 	imes \frac{R_2}{R_1 + R_2}$Here, $V_{total} = 60 \, V$, $R_1 = 64 \, \Omega$, and $R_2 = 32 \, \Omega$.$V_J = 60 	imes \frac{32}{64 + 32}$$V_J = 60 	imes \frac{32}{96}$$V_J = 60 	imes \frac{1}{3}$$V_J = 20 \, V$

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

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