The current in the circuit will be (a) 5/40 A (b) 5/50 A (c) 5/10 A (d) 5/20 A

Answer: (b). The battery sends current through the lower 20 Ω to the left junction, then left to right through the branches. D₁ points left, so it is reverse biased and the top branch is open. D₂ points right, so it is forward biased and the 30 Ω branch conducts. I = (5)/(20 + 30) = 5/50 A So the answer is (b) 5/50 A.

The current in the circuit will be | NEET 2001 (AIPMT)

Q121NEET 2001 · AIPMTMedium

P-N Junction Diode

The current in the circuit will be

Show answer and solution

Answer: (b)

The battery sends current through the lower $20\ \Omega$ to the left junction, then left to right through the branches.
$D_1$ points left, so it is reverse biased and the top branch is open.
$D_2$ points right, so it is forward biased and the $30\ \Omega$ branch conducts.
$I = \dfrac{5}{20 + 30} = \dfrac{5}{50}$ A
So the answer is (b) $5/50$ A.

Report a mistake

P-N Junction Diode question from NEET 2001 (AIPMT)

More questions on P-N Junction Diode