In the figure, potential difference between A and B is: (a) 10 V (b) 5 V (c) zero (d) 15 V

Answer: (a). The diode is forward biased (ideal, short), so the two 10 kΩ resistors from A to B are in parallel: 5 kΩ. Total = 10 + 5 = 15 kΩ, so I = (30)/(15 kΩ) = 2 mA VAB = 2× 10⁻³× 5× 10³ = 10 V So the answer is (a) 10 V.

In the figure, potential difference between A and B is: | JEE Main 2020 (7 Jan, Shift 2)

Q84JEE Main 2020 · 7 JAN · SHIFT 2Medium

P-N Junction Diode

In the figure, potential difference between A and B is:

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Answer: (a)

The diode is forward biased (ideal, short), so the two $10\text{ k}\Omega$ resistors from A to B are in parallel: $5\text{ k}\Omega$ .
Total $= 10 + 5 = 15\text{ k}\Omega$ , so $I = \dfrac{30}{15\text{ k}\Omega} = 2$ mA
$V_{AB} = 2\times 10^{-3}\times 5\times 10^{3} = 10$ V
So the answer is (a) $10$ V.

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