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In the circuit shown in figure, if the diode forward voltage is $0.3 \mathrm{~V}$, the voltage difference between $A$ and $B$ is
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The correct answer is:
$2.3 \mathrm{~V}$
Let $V$ be the potential difference between $A$ and $B$, then
$\begin{array}{ll} & V-0.3=(5+5) \times 10^3 \times\left(0.2 \times 10^{-3}\right)=2 \\ \text { or } & V=2+0.3=2.3 \mathrm{~V}\end{array}$
$\begin{array}{ll} & V-0.3=(5+5) \times 10^3 \times\left(0.2 \times 10^{-3}\right)=2 \\ \text { or } & V=2+0.3=2.3 \mathrm{~V}\end{array}$
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