The circuit shown below contains two ideal diodes, each with a forward resistance of 50 Ω. If the battery voltage is 6 V, the current through the 100 Ω resistance (in Amperes) is: (a) 0.027 (b) 0.030 (c) 0.036 (d) 0.020

Answer: (d). The left terminal of the battery is positive. D₁ points right (forward, 50 Ω); D₂ points left (reverse, open). Total resistance = 100 + 150 + 50 = 300 Ω I = 6/300 = 0.02 A So the answer is (d) 0.020 A.

The circuit shown below contains two ideal diodes, each with a… | JEE Main 2019 (11 Jan, Shift 2)

Q89JEE Main 2019 · 11 JAN · SHIFT 2Medium

P-N Junction Diode

The circuit shown below contains two ideal diodes, each with a forward resistance of 50 $\Omega$ . If the battery voltage is 6 V, the current through the 100 $\Omega$ resistance (in Amperes) is:

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Answer: (d)

The left terminal of the battery is positive. $D_1$ points right (forward, $50\ \Omega$ ); $D_2$ points left (reverse, open).
Total resistance $= 100 + 150 + 50 = 300\ \Omega$
$I = \dfrac{6}{300} = 0.02$ A
So the answer is (d) $0.020$ A.

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P-N Junction Diode question from JEE Main 2019 (11 Jan, Shift 2)

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