MHTCET 19 Apr 2026 Shift 2 Paper

Concept:The induced emf in the secondary coil depends on mutual inductance and the rate of change of primary current. Then the maximum current is found using the coil's impedance.Explanation:Given primary current is $I_{p}=12\sin(50\pi t)$, so the angular frequency is $\omega=50\pi\ \text{rad s}^{-1}$.The maximum induced emf in the secondary is given by $E_{\max}=M\omega I_{p,\max}$.Here, $M=\pi\ \text{mH}=\pi\times10^{-3}\ \text{H}$ and $I_{p,\max}=12\ \text{A}$.Thus, $E_{\max}=(\pi\times10^{-3})(50\pi)(12)=0.6\pi^{2}$.Using $\pi^{2}=10$, we get $E_{\max}=0.6\times10=6\ \text{V}$.The secondary coil has self-inductance $L=\frac{60}{\pi}\ \text{mH}=\frac{60}{\pi}\times10^{-3}\ \text{H}$.Its inductive reactance is $X_{L}=\omega L=(50\pi)\left(\frac{60}{\pi}\times10^{-3}\right)=3\ \Omega$.The coil's resistance is $R=4\ \Omega$, so impedance is $Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{4^{2}+3^{2}}=5\ \Omega$.Hence, the maximum induced current in coil S is $I_{\max}=\frac{E_{\max}}{Z}=\frac{6}{5}=1.2\ \text{A}$.Answer:The correct option is D, i.e., $1.2\ \text{A}$.