The current i in the network is: (a) 0.6 A (b) 0.3 A (c) 0 A (d) 0.2 A

Answer: (b). For the battery polarity shown, both diodes are reverse biased, so their branches ( 5 Ω top-left and 20 Ω bottom-right) carry no current. The remaining path is 10 Ω (left to bottom), 5 Ω (bottom to top), 10 Ω (top to right), all in series: 25 Ω. Total with the external 5 Ω: R = 30 Ω i = 9/30 = 0.3 A So the answer is (b) 0.3 A.

The current i in the network is: | JEE Main 2020 (9 Jan, Shift 2)

Q83JEE Main 2020 · 9 JAN · SHIFT 2Medium

P-N Junction Diode

The current i in the network is:

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Answer: (b)

For the battery polarity shown, both diodes are reverse biased, so their branches ( $5\ \Omega$ top-left and $20\ \Omega$ bottom-right) carry no current.
The remaining path is $10\ \Omega$ (left to bottom), $5\ \Omega$ (bottom to top), $10\ \Omega$ (top to right), all in series: $25\ \Omega$ .
Total with the external $5\ \Omega$ : $R = 30\ \Omega$
$i = \dfrac{9}{30} = 0.3$ A
So the answer is (b) $0.3$ A.

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P-N Junction Diode question from JEE Main 2020 (9 Jan, Shift 2)

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