MCQ
In the given circuit, the potential of the point $E$ is
  • A
    Zero
  • B
    $ - \,8\,V$
  • $ - \,4/3\,V$
  • D
    $4/3 \,V$

Answer

Correct option: C.
$ - \,4/3\,V$
c
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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