In the given figure, each diode has a forward bias resistance of 30 Ω and infinite resistance in reverse bias. The current I1 will be: (a) 3.75 A (b) 2.35 A (c) 2 A (d) 2.73 A

Answer: (c). The top two diodes are forward biased; the third is reverse biased (open). Each conducting branch: 30 + 130 = 160 Ω; two in parallel give 80 Ω. Total = 80 + 20 = 100 Ω I₁ = (200)/(100) = 2 A So the answer is (c) 2 A.

In the given figure, each diode has a forward bias resistance of 30 Ω… | JEE Main 2021 (1 Sep, Shift 2)

Q75JEE Main 2021 · 1 SEP · SHIFT 2Medium

P-N Junction Diode

In the given figure, each diode has a forward bias resistance of 30 $\Omega$ and infinite resistance in reverse bias. The current I1 will be:

Show answer and solution

Answer: (c)

The top two diodes are forward biased; the third is reverse biased (open).
Each conducting branch: $30 + 130 = 160\ \Omega$ ; two in parallel give $80\ \Omega$ .
Total $= 80 + 20 = 100\ \Omega$
$I_1 = \dfrac{200}{100} = 2$ A
So the answer is (c) $2$ A.

Report a mistake

P-N Junction Diode question from JEE Main 2021 (1 Sep, Shift 2)

More questions on P-N Junction Diode