Question:

All the elements in the circuit shown in the following figure are ideal. Which of the following statements is/are true? \begin{center} \includegraphics[width=0.65\textwidth]{21.jpeg} \end{center}

Show Hint

When analyzing diode circuits, always compare node voltages. A diode conducts if its anode is at higher potential than its cathode (by at least the threshold, here ideal so 0V).
Updated On: Aug 28, 2025
  • 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
Hide Solution
collegedunia
Verified By Collegedunia

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:
\[ \boxed{(B) \; \text{and} \; (C)} \]

Was this answer helpful?
0
0

Questions Asked in GATE EE exam

View More Questions