Question

In the circuit shown in figure. If the diode forward voltage drop is $0.3V$, the voltage difference between $A$ and $B$ is

• A. 1.3 V
• B. 2.3 V
• C. 0
• D. 0.5 V

Answer 1

Answer: (B)

Let $V$ be the potential difference between $A$ and $B$.
Then $V-0.3=(5+5)\times 10^3\times (0.2\times 10^{-3})=2$
or $V=2+0.3=2.3V$