In the given circuit, the potential of the point $E$ is
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The current in the circuit $ = \frac{8}{{5 + 1}} = \frac{4}{3}$
Now ${V_C} - {V_E} = \frac{4}{3} \times 1$ $ \Rightarrow $ ${V_E} = - \frac{4}{3}\,V$
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