Two coils P and S have a mutual inductance of ( π) mH. The secondary coil S has resistance 4 Ω and self inductance (60/π) mH. If the current in the primary is Ip = 12sin (50π t), then the maximum value of the current induced in coil S is [Take π² = 10 ] (a) 2 A (b) 1.8 A (c) 1.5 A (d) 1.2 A

Answer: (d). Induced emf in S: e = M(dIp)/(dt), so emax = M I₀ ω = (π×10⁻³)(12)(50π) = 0.6π² = 6 V Reactance of S: XL = ω Ls = 50π × 60/π×10⁻³ = 3 Ω Impedance of S: Z = √(4² + 3²) = 5 Ω Imax = 6/5 = 1.2 A So the answer is (d) 1.2 A.

Two coils P and S have a mutual inductance of ( π) mH. The secondary… | MHT CET 2026 (19 Apr, Shift 2)

Q1MHT CET 2026 · 19 APR · SHIFT 2Medium

Inductors & DC Transients (Mixed)

Two coils P and S have a mutual inductance of ( $\pi$ ) mH. The secondary coil S has resistance $4\ \Omega$ and self inductance $(60/\pi)$ mH. If the current in the primary is $I_p = 12\sin(50\pi t)$ , then the maximum value of the current induced in coil S is [Take $\pi^2 = 10$ ]

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Answer: (d)

Induced emf in S: $e = M\frac{dI_p}{dt}$ , so $e_{max} = M I_0 \omega = (\pi\times10^{-3})(12)(50\pi) = 0.6\pi^2 = 6\text{ V}$
Reactance of S: $X_L = \omega L_s = 50\pi \times \frac{60}{\pi}\times10^{-3} = 3\ \Omega$
Impedance of S: $Z = \sqrt{4^2 + 3^2} = 5\ \Omega$
$I_{max} = \frac{6}{5} = 1.2\text{ A}$
So the answer is (d) 1.2 A.

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