All the elements in the circuit shown in the following figure are ideal. Which of the following statements is/are true?
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Show Hint
- When switch \(S\) is ON, both \(D_1\) and \(D_2\) conduct and \(D_3\) is reverse biased
- When switch \(S\) is ON, \(D_1\) conducts and both \(D_2\) and \(D_3\) are reverse biased
- When switch \(S\) is OFF, \(D_1\) is reverse biased and both \(D_2\) and \(D_3\) conduct
- When switch \(S\) is OFF, \(D_1\) conducts, \(D_2\) is reverse biased and \(D_3\) conducts
The Correct Option is B, C
Solution and Explanation
Step 1: Case when switch \(S\) is ON.
- The 2A current source is directly connected across the 20V source.
- The voltage at the junction tends to rise such that diode \(D_1\) becomes forward biased.
- Diode \(D_2\) sees higher potential at its cathode (20V) compared to anode (10V), hence reverse biased.
- Diode \(D_3\) sees higher potential at its cathode (40V) than anode (20V), hence reverse biased.
Thus, when \(S\) is ON:
\[
D_1 \, \text{conducts}, D_2 \, \text{OFF}, D_3 \, \text{OFF}
\]
This matches option (B).
Step 2: Case when switch \(S\) is OFF.
- The 2A current source is no longer shorted; it tries to push current.
- \(D_1\) has 20V at its anode and higher potential at cathode side due to current source, so it becomes reverse biased.
- Current from 10V source and 20V source flows through \(D_2\) (anode at 10V, cathode at 20V → forward biased).
- Similarly, path through \(D_3\) (anode at 20V, cathode at 40V → forward biased).
Thus, when \(S\) is OFF:
\[
D_1 \, \text{OFF}, D_2 \, \text{ON}, D_3 \, \text{ON}
\]
This matches option (C).
Step 3: Eliminate wrong options.
- (A) claims \(D_1\) and \(D_2\) conduct with \(S\) ON — incorrect since \(D_2\) is reverse biased.
- (D) claims \(D_1\) and \(D_3\) conduct with \(S\) OFF — incorrect since \(D_1\) is OFF.
Final Answer:
\[
\boxed{(B) \; \text{and} \; (C)}
\]
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