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# In the circuit shown in the figure , if the diode forward voltage is $0.3\;V$ , the voltage difference between A and B is

$\begin{array}{1 1} 1.3\;V \\ 2.3\;V \\ 0 \\ 0.5\;V \end{array}$

$V-0.3 = (5+5) \times 10^3 \times (0.2 \times 10^{-3})^2=2$
$V= 2+0.3 =2.3\;V$