MCQ
In the circuit shown in Fig., if the diode forward voltage drop is $0.3V,$ the voltage difference between $A$ and $B$ is: 
- A$1.3V.$
- ✓$2.3V.$
- C$0.$
- D$0.5V.$
✓
Answer
Correct option: B.
$2.3V.$
If $V$ is the potential difference between $A$ and $B$, then according to the questions and using Kirchhoff’s law, we have
$V-0.3=\left[(5+5) 10^3\right] \times\left(0.2 \times 1\left(10^{-3}\right)\right)$
$\Rightarrow V-0.3=\left[(5+5) 10^3\right] \times\left(0.2 \times 10^{-3}\right)$
$\Rightarrow V-0.3=10 \times 10^3 \times 0.2 \times 10^{-3}=2$
$\Rightarrow V=2+0.3=2.3 V$
$V-0.3=\left[(5+5) 10^3\right] \times\left(0.2 \times 1\left(10^{-3}\right)\right)$
$\Rightarrow V-0.3=\left[(5+5) 10^3\right] \times\left(0.2 \times 10^{-3}\right)$
$\Rightarrow V-0.3=10 \times 10^3 \times 0.2 \times 10^{-3}=2$
$\Rightarrow V=2+0.3=2.3 V$
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