Two circuits A and B are connected to identical d.c. sources each of e.m.f. 10 volt. Self-inductances of circuits A and B are respectively LA = 10 H and LB = 10 mH. The total resistance of each circuit is 40 Ω. The ratio of energy consumed in circuit A and circuit B to build up the current to steady value is (a) 800 (b) 1000 (c) 1200 (d) 1400

Answer: (b). Steady current in both: I = E/R = 10/40 = 0.25 A. Energy stored in building up the current: U = 1/2L I², so U ∝ L for the same I. (UA)/(UB) = 1010× 10⁻³ = 1000 So the answer is (b) 1000.

Two circuits A and B are connected to identical d.c. sources each of… | MHT CET 2025 (26 Apr, Shift 1)

Q31MHT CET 2025 · 26 APR · SHIFT 1Medium

Self-Inductance & Energy

Two circuits A and B are connected to identical d.c. sources each of e.m.f. 10 volt. Self-inductances of circuits A and B are respectively $L_A = 10\text{ H}$ and $L_B = 10\text{ mH}$ . The total resistance of each circuit is $40\ \Omega$ . The ratio of energy consumed in circuit A and circuit B to build up the current to steady value is

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Answer: (b)

Steady current in both: $I = \dfrac{E}{R} = \dfrac{10}{40} = 0.25$ A.
Energy stored in building up the current: $U = \dfrac{1}{2}L I^2$ , so $U \propto L$ for the same $I$ .
$\dfrac{U_A}{U_B} = \dfrac{10}{10\times 10^{-3}} = 1000$
So the answer is (b) 1000.

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