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In the figure shown, if the diode forward voltage drop is \( 0.2 \mathrm{~V} \), the voltage difference between \( \mathrm{A} \)
and \( \mathrm{B} \) is
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and \( \mathrm{B} \) is
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Verified Answer
The correct answer is:
\( 2.2 \mathrm{~V} \)
Given $I=0.2 \mathrm{~mA}=0.2 \times 10^{-3} \mathrm{~A} ;$ Voltage drop at diode $=0.2 \mathrm{~V} R_{1}=R_{2}=5 \mathrm{k} \Omega=5 \times 10^{3} \Omega$
Now, voltage drop at $R_{1}=0.2 \times 10^{-3} \times 5 \times 10^{3}=1 \mathrm{~V}$
and voltage drop at $R_{2}=0.2 \times 10^{-3} \times 5 \times 10^{3}=1 \mathrm{~V}$
Therefore, voltage difference $=1+1+0.2=2.2 \mathrm{~V}$
Now, voltage drop at $R_{1}=0.2 \times 10^{-3} \times 5 \times 10^{3}=1 \mathrm{~V}$
and voltage drop at $R_{2}=0.2 \times 10^{-3} \times 5 \times 10^{3}=1 \mathrm{~V}$
Therefore, voltage difference $=1+1+0.2=2.2 \mathrm{~V}$
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