In the given circuit, the current (I) through the battery will be (a) 1A (b) 2.5 A (c) 2 A (d) 1.5 A

Answer: (d). Call the left node L, top-right T and bottom-right B. Current leaves the battery at B and returns at L. D₃ (B to L) is forward biased: one 10 Ω path. D₁ (T to L) is forward biased, so B to T ( 10 Ω) to L ( 10 Ω) is a 20 Ω path. D₂ (L to T) is reverse biased. Req = (10× 20)/(10 + 20) = 20/3 Ω I = (10)/(20/3) = 1.5 A So the answer is (d) 1.5 A.

In the given circuit, the current (I) through the battery will be | JEE Main 2023 (15 Apr, Shift 1)

Q58JEE Main 2023 · 15 APR · SHIFT 1Medium

P-N Junction Diode

In the given circuit, the current (I) through the battery will be

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Answer: (d)

Call the left node L, top-right T and bottom-right B. Current leaves the battery at B and returns at L.
$D_3$ (B to L) is forward biased: one $10\ \Omega$ path. $D_1$ (T to L) is forward biased, so B to T ( $10\ \Omega$ ) to L ( $10\ \Omega$ ) is a $20\ \Omega$ path. $D_2$ (L to T) is reverse biased.
$R_{eq} = \dfrac{10\times 20}{10 + 20} = \dfrac{20}{3}\ \Omega$
$I = \dfrac{10}{20/3} = 1.5$ A
So the answer is (d) $1.5$ A.

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P-N Junction Diode question from JEE Main 2023 (15 Apr, Shift 1)

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