Explanation

(c) In the given circuit, diode $D_{1}$ is reverse biased, so it will not conduct. Diodes $D_{2}$ and $D_{3}$ are forward biased, so they will conduct. The corresponding equivalent circuit is as shown in the figure
[image:14_image208.emf]
The equivalent resistance of the circuit is
$R_{eq} = \frac{(5 + 5) \times 20}{(5 + 5) + 20} = \frac{10 \times 20}{10 + 20} = \frac{200}{30} = \frac{20}{3}\Omega$
Current through the battery, $I = \frac{10V}{\frac{20}{3}\Omega} = 1.5A$