Explanation

(c) Before connecting $E,$ the circuit diagram is as shown in the figure
[image:3_image186.emf]
The equivalent resistance of the given circuit is
$R_{eq} = 6\ \Omega + 8\ \Omega + 10\ \Omega = 24\ \Omega$
Current in the circuit, $I = \frac{12V}{24\Omega} = \frac{1}{2}A$
Before connecting $E,$ the current through $8\ \Omega$ is
$I = \frac{1}{2}A$
After connecting $E,$ the current through $8\ \Omega$ is also
$I = \frac{1}{2}A$
$\therefore E = \frac{1}{2}A \times 8\Omega = 4V$