NEET 2026 — Physics solution

NEET2026Physics

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

A
$\frac{1}{3}$ A
B
$\frac{15}{2}$ A
C
$\frac{5}{3}$ A
D
$\frac{5}{9}$ A

Solution

Diodes $D_{1}$ and $D_{3}$ are forward biased (closed switches); $D_{2}$ and $D_{4}$ are reverse biased (open). So only the 4 $Ω$ and 2 $Ω$ branches conduct.

These two branches are in parallel:

$R_{e q} = \frac{4 \times 2}{4 + 2} = \frac{4}{3} Ω$

$I = \frac{V}{R _{e q}} = \frac{10}{4/3} = \frac{15}{2}$ A.

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