In the circuit shown in figure all the three diodes D1, D2, D3 have forward resistance 50 Ω each and infinite backward resistance. If the battery voltage is 5V, the current through 100 Ω resistance is (a) 60 mA (b) 30 mA (c) 20 mA (d) 10 mA

Answer: (b). The left terminal of the battery is positive. D₁ and D₂ point right (forward), D₃ points left (reverse, open). Branch resistances: D₁ branch = 50 + 150 = 200 Ω, D₂ branch = 50 + 50 = 100 Ω. Parallel: (200× 100)/(300) = 200/3 Ω; total = 100 + 200/3 = 500/3 Ω I = (5)/(500/3) = 0.03 A = 30 mA through the 100 Ω resistor. So the answer is (b) 30 mA.

In the circuit shown in figure all the three diodes D1, D2, D3 have… | MHT CET 2026 (17 Apr, Shift 1)

Q15MHT CET 2026 · 17 APR · SHIFT 1Medium

P-N Junction Diode

In the circuit shown in figure all the three diodes D1, D2, D3 have forward resistance $50\ \Omega$ each and infinite backward resistance. If the battery voltage is 5V, the current through $100\ \Omega$ resistance is

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Answer: (b)

The left terminal of the battery is positive. $D_1$ and $D_2$ point right (forward), $D_3$ points left (reverse, open).
Branch resistances: $D_1$ branch $= 50 + 150 = 200\ \Omega$ , $D_2$ branch $= 50 + 50 = 100\ \Omega$ .
Parallel: $\dfrac{200\times 100}{300} = \dfrac{200}{3}\ \Omega$ ; total $= 100 + \dfrac{200}{3} = \dfrac{500}{3}\ \Omega$
$I = \dfrac{5}{500/3} = 0.03$ A $= 30$ mA through the $100\ \Omega$ resistor.
So the answer is (b) $30$ mA.

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