Explanation

(c) Net voltage =12-8=4V
Net resistance =9+2+1=12$\Omega$
Current through the circuit
$= \frac{4}{12} = \frac{1}{3}A$
Terminal voltage across $E_{2}$ is
$= E_{2} - Ir_{2} = 12 - \frac{1}{3} \times 2$
$= 12 - \frac{2}{3} = \frac{34}{3} = 11.34\ V$