In the following circuit shown in figure, three diodes are connected each with forward resistance 40 Ω and infinite backward resistance. The current through 100 Ω resistance is (a) 18 mA (b) 36 mA (c) 9 mA (d) 27 mA

Answer: (b). The top rail is at the battery's positive terminal. D₁ points up (reverse, open); D₂ and D₃ point down (forward). Branches: D₂: 40 + 60 = 100 Ω; D₃: 40 + 160 = 200 Ω. Parallel = (100× 200)/(300) = 200/3 Ω. Total = 100 + 200/3 = 500/3 Ω I = (6)/(500/3) = 0.036 A = 36 mA So the answer is (b) 36 mA.

In the following circuit shown in figure, three diodes are connected… | MHT CET 2025 (20 Apr, Shift 2)

Q32MHT CET 2025 · 20 APR · SHIFT 2Medium

P-N Junction Diode

In the following circuit shown in figure, three diodes are connected each with forward resistance $40\ \Omega$ and infinite backward resistance. The current through $100\ \Omega$ resistance is

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Answer: (b)

The top rail is at the battery's positive terminal. $D_1$ points up (reverse, open); $D_2$ and $D_3$ point down (forward).
Branches: $D_2$ : $40 + 60 = 100\ \Omega$ ; $D_3$ : $40 + 160 = 200\ \Omega$ . Parallel $= \dfrac{100\times 200}{300} = \dfrac{200}{3}\ \Omega$ .
Total $= 100 + \dfrac{200}{3} = \dfrac{500}{3}\ \Omega$
$I = \dfrac{6}{500/3} = 0.036$ A $= 36$ mA
So the answer is (b) $36$ mA.

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