Q 12-03-006NEETNEET 2025Top questionMedium
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 :
Answer: (B) $2.0$ A
C and D are joined by a wire, so they are at the same potential. The circuit is ($1\ \Omega \parallel 3\ \Omega$) in series with ($2\ \Omega \parallel 4\ \Omega$):
$$R = \frac{1 \times 3}{4} + \frac{2 \times 4}{6} = \frac{3}{4} + \frac{4}{3} = \frac{25}{12}\ \Omega$$
$$I = \frac{50}{25/12} = 24\ \text{A}$$
Current through $1\ \Omega$ (into C): $24 \times \dfrac{3}{4} = 18$ A.
Current through $2\ \Omega$ (out of C): $24 \times \dfrac{4}{6} = 16$ A.
Current in CD $= 18 - 16 = 2.0$ A (from C to D).
Solution by Sreeraj P, M.Sc Physics