NEET UGPhysicsModerateNEET 2025

A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is:

[IMAGE]

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Sol.
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R_{AB} = (1\Omega/3\Omega)

in series with

(2\Omega/4\Omega)

= \frac{3 \times 1}{3+1} + \frac{2 \times 4}{2+4}

= \frac{3}{4} + \frac{8}{6} = \frac{9+16}{12} = \frac{25}{12} \Omega

Now total current through cell

I = \frac{50}{25/12} = 24A

I_{1\Omega} = \frac{3}{4} \times 24 = 18A, I_{3\Omega} = \frac{1}{4} \times 24 = 6A

I_{2\Omega} = \frac{4}{6} \times 24 = 16A, I_{4\Omega} = \frac{2}{6} \times 24 = 8A

Using junction rule at C,

I_{CD} = 18 - 16 = 2A

(From C to D)

Source: geekyrakshit/indian-exam-questions / neet2025-code45

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