semiconductor-electronics--materials-devices-and-simple-circuits Question 7
Question: Q. 10. In the circuit shown in Figure, if the diode forward voltage drop is $0.3 \mathrm{~V}$, the voltage difference between $A$ and $B$ is
(a) $1.3 \mathrm{~V}$
(b) $2.3 \mathrm{~V}$
(c) $0 \mathrm{~V}$
(d) $0.5 \mathrm{~V}$
[NCERT Exemplar]
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Solution:
Ans. Correct option : (b)
Explanation: In the middle right of the circuit the capacitor behaves like an open circuit for $d c 0.2 \mathrm{~mA}$ current so current will flow from $A$ to $B$ only. Let potential across $A$ and $B$ is $V$, so by
Kirchhoff’s loop law,
$V_{A B}=\left(5000 \times 0.2 \times 10^{-3}\right)+0.3+5000 \times 0.2 \times 10^{-3}$
$V_{A B}=1 \mathrm{~V}+0.3 \mathrm{~V}+1 \mathrm{~V}$
$V_{A B}=2.3 \mathrm{~V}$
