JEE Main 2025 Jan 23 Shift 2, Physics Q16: Semiconductors, p-n Junction Diode and Applications

JEE Main2025Jan 23, Shift 2Physics

What is the current through the battery in the circuit shown below?

Solution

With the polarity shown, both diodes are forward biased and behave as short circuits (zero forward drop, ideal).

The two $20 Ω$ resistors are then in parallel: $R_{e q} = \frac{20 \cdot 20}{20 + 20} = 10 Ω.$

Battery current: $i = \frac{V}{R _{e q}} = \frac{5}{10} = 0.5 A .$

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