KCET · Physics · Current Electricity
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
- A \( 1.3 \mathrm{~V} \)
- B \( 2.2 \mathrm{~V} \)
- C \( 00 \)
- D \( 0.5 \mathrm{~V} \)
Answer & Solution
Correct Answer
(B) \( 2.2 \mathrm{~V} \)
Step-by-step Solution
Detailed explanation
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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