Question

The circuit has two oppositely connected ideal diodes in parallel. The current flowing in the circuit will be

• A. \(2\,\text{A}\)
• B. \(1.71\,\text{A}\)
• C. \(2.31\,\text{A}\)
• D. \(1.33\,\text{A}\)

Hint:

For ideal diodes: \[ \text{Forward biased diode} \Rightarrow \text{Short circuit} \] \[ \text{Reverse biased diode} \Rightarrow \text{Open circuit} \] Always replace the diodes first and then solve the simplified resistive circuit.

Answer 1

The Correct Option is B

Concept: An ideal diode offers: \[ R_f=0 \] when forward biased and \[ R_r=\infty \] when reverse biased. Therefore, only the branch containing the forward-biased diode conducts current.

Step 1:Identify the conducting diode. The upper node is connected to the positive terminal of the battery. Hence diode \(D_1\) is forward biased and diode \(D_2\) is reverse biased. Therefore, \[ D_1 \text{ conducts} \] and \[ D_2 \text{ does not conduct.} \]

Step 2: Redraw the equivalent circuit. Since \(D_1\) is ideal and forward biased, it behaves as a short circuit. Since \(D_2\) is reverse biased, it behaves as an open circuit. The circuit reduces to a series combination of \[ 4\Omega \] and \[ 3\Omega \] resistors. \[ R_{\text{eq}}=4+3=7\Omega \]

Step 3: Calculate the current. Applying Ohm's law, \[ I=\frac{V}{R_{\text{eq}}} \] \[ I=\frac{12}{7} \] \[ I=1.714\,\text{A} \] \[ I\approx1.71\,\text{A} \]

Step 4: State the answer. \[ { I=1.71\,\text{A} } \] Hence, the correct option is \[ {(B)} \]