Coupled Circuits Questions and Answers

Q: The current in a coil decreases from 10 A to 4 A in 0.1 second. If the inductance of the coil is 4 H, find the emf induced in the coil.

The induced emf is 240 V.The emf of self-induction is the inductance times the rate of change of current, \(e = L\,\frac{\Delta I}{\Delta t}\).\[ \begin{aligned} \Delta I &= 10 - 4 = 6\ \text{A} \\ e &= 4 \times \frac{6}{0.1} = 240\ \text{V} \end{aligned} \]Using the final current of 4 A instead of the change gives 160 V, and using the initial 10 A gives 400 V. Only the change in current induces an emf.

Two coils have self-inductance of 20 H and 20 H connected in series and mutual inductance between two coils is 2.5 H. The value of total equivalent inductance is:

A.42.5 H
B.45 H
C.40 H
D.12.5 H

Exam Year(s)

DSSSB AE - 2023,

Answer: B

When two coils are connected in series so that their fluxes aid, the total inductance is the sum of the self inductances plus twice the mutual inductance.

\[ \begin{aligned} L = L_1 + L_2 + 2M = 20 + 20 + 2 \times 2.5 \\ &= 45\ \text{H} \end{aligned} \]

The item does not say that the coils oppose, so the aiding connection is taken, the usual default. With opposing fluxes the result would be 35 H, which is not offered. The value 42.5 H would be \(L_1 + L_2 + M\), and 40 H ignores the coupling. The answer is 45 H, option B.

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When two coils having inductance of L₁ and L₂ are connected through mutual inductance M, the coefficient of coupling K is given by :

A.\(K = \frac{M}{\sqrt{L_{1}L_{2}}}\)
B.\(K = \frac{M}{\sqrt{2L_{1}L_{2}}}\)
C.\(K = \frac{L_{1}L_{2}}{M}\)
D.\(K = \frac{2M}{\sqrt{L_{1}L_{2}}}\)

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: A

The coefficient of coupling is \(K = \frac{M}{\sqrt{L_{1}L_{2}}}\).

The mutual inductance between two coils can never exceed the geometric mean of their self-inductances; it reaches \(\sqrt{L_{1}L_{2}}\) only when every line of flux from one coil links the other. K measures how much of that ideal is achieved, so it is the ratio of M to \(\sqrt{L_{1}L_{2}}\) and always lies between 0 and 1.

The factor 2 in the other two square-root options is wrong: the version with 2M would give K = 2 for perfect coupling, which is impossible. L₁L₂/M has the units of inductance, not a pure number.

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The current in a coil decreases from 10 A to 4 A in 0.1 second. If the inductance of the coil is 4 H, find the emf induced in the coil.

A.230 V
B.400 V
C.240 V
D.160 V

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: C

The induced emf is 240 V.

The emf of self-induction is the inductance times the rate of change of current, \(e = L\,\frac{\Delta I}{\Delta t}\).

\[ \begin{aligned} \Delta I &= 10 - 4 = 6\ \text{A} \\ e &= 4 \times \frac{6}{0.1} = 240\ \text{V} \end{aligned} \]

Using the final current of 4 A instead of the change gives 160 V, and using the initial 10 A gives 400 V. Only the change in current induces an emf.

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