Question:
The circuit shown below contains two ideal diodes, each with a forward resistance of \( 50\ \Omega \). If the battery voltage is 6 V, the current through the \( 100\ \Omega \) resistance (in Amperes) is : 
The circuit shown below contains two ideal diodes, each with a forward resistance of \( 50\ \Omega \). If the battery voltage is 6 V, the current through the \( 100\ \Omega \) resistance (in Amperes) is :
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Check which diode is forward biased from the battery polarity; the reverse-biased branch carries no current.
Updated On: Nov 5, 2026
- 0.027
- 0.030
- 0.036
- 0.020
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The Correct Option is D
Solution and Explanation
The longer plate (positive terminal) of the 6 V battery connects to the left rail, which is the anode side of D1 and the cathode side of D2.
So D1 is forward biased (conducts, 50 \( \Omega \)) and D2 is reverse biased (ideal: no current), so the 75 \( \Omega \) branch is inactive.
Total resistance in the loop: \( 50 + 150 + 100 = 300\ \Omega \).
\( I = \frac{6}{300} = 0.02\ \mathrm{A} \).
Answer: (4) 0.020
So D1 is forward biased (conducts, 50 \( \Omega \)) and D2 is reverse biased (ideal: no current), so the 75 \( \Omega \) branch is inactive.
Total resistance in the loop: \( 50 + 150 + 100 = 300\ \Omega \).
\( I = \frac{6}{300} = 0.02\ \mathrm{A} \).
Answer: (4) 0.020
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