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