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The current I in the circuit shown is: (All diodes are ideal and identical)

Asked in NEET 2026 · Circuits with diodes

• (1) 5/3 A
• (2) 5/9 A
• (3) 1/3 A
• (4) (15)/2 A

Step-by-step solution

Why the other options are wrong

• (1) 5/3 A is (10)/6, the two conducting branches added in series as 4+2=6 Ω. They sit side by side across the same battery, so their currents add instead of their resistances.
• (2) comes from no single slip here produces 5/9 A. As a check: 4 Ω and 5 Ω in series would carry (10)/9 A, twice this figure, and the 5 Ω branch is blocked in any case.
• (3) comes from no single slip here produces 1/3 A either. As a check: the least any single branch of this network can carry is (10)/5=2 A, so a third of an amp is far below anything on offer.

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