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(D) The circuit consists of an $8 \, V$ battery,a $5 \, \Omega$ resistor,and a $1 \, \Omega$ resistor in series. The total resistance $R_{eq} = 5 \, \

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$4/3 \, V$

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(D) The circuit consists of an $8 \, V$ battery,a $5 \, \Omega$ resistor,and a $1 \, \Omega$ resistor in series. The total resistance $R_{eq} = 5 \, \Omega + 1 \, \Omega = 6 \, \Omega$.The current $I$ in the circuit is given by $I = \frac{V}{R_{eq}} = \frac{8 \, V}{6 \, \Omega} = \frac{4}{3} \, A$.Point $C$ is grounded,so its potential $V_C = 0 \, V$.Moving from $C$ to $E$ through the $1 \, \Omega$ resistor,the potential increases because the current flows from $E$ to $D$ to $C$. Thus,$V_E - V_C = I 	imes R = \frac{4}{3} \, A 	imes 1 \, \Omega = \frac{4}{3} \, V$.Since $V_C = 0 \, V$,we have $V_E = \frac{4}{3} \, V$.

In the given circuit,the potential of the point $E$ is

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