Question

In the given circuit, the current through the battery is :

• A. 1. 5 A
• B. 2 A
• C. 3 A
• D. 5.33 A

Answer 1

The correct option is C 3 A

In the given circuit diode $D_2$ is reverse biased, so it will not conduct. Diodes $D_1$ and $D_3$ are forward biased, so they will conduct. Thus the equivalent circuit is
$\therefore R_{net}=\frac{(4+4)\times 16}{(4+4)+16}=\frac{8\times 16}{24}=\frac{16}{3}\Omega$
$\therefore$ Current through battery,
$I=\frac{V}{R_{net}}=\frac{16}{16/3}=3A$