Two coils P and S have a mutual inductance of | vedclass

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(D) The induced electromotive force $(EMF)$ in the secondary coil is given by $\varepsilon = M \frac{dI_p}{dt}$.Given $I_p = 12 \sin(50\pi t)$, then $

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$1.2$

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(D) The induced electromotive force $(EMF)$ in the secondary coil is given by $\varepsilon = M \frac{dI_p}{dt}$.Given $I_p = 12 \sin(50\pi t)$, then $\frac{dI_p}{dt} = 12 	imes 50\pi \cos(50\pi t) = 600\pi \cos(50\pi t)$.The maximum induced $EMF$ is $\varepsilon_{max} = M 	imes (600\pi) = (\pi 	imes 10^{-3}) 	imes 600\pi = 600\pi^2 	imes 10^{-3} 	ext{ V}$.Using $\pi^2 = 10$, we get $\varepsilon_{max} = 600 	imes 10 	imes 10^{-3} = 6 	ext{ V}$.The impedance $Z$ of the secondary coil is $Z = \sqrt{R^2 + X_L^2}$, where $X_L = \omega L$.Here $\omega = 50\pi 	ext{ rad/s}$ and $L = (60/\pi) 	imes 10^{-3} 	ext{ H}$.$X_L = 50\pi 	imes \frac{60}{\pi} 	imes 10^{-3} = 3000 	imes 10^{-3} = 3 	ext{ } \Omega$.$Z = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 	ext{ } \Omega$.The maximum induced current $I_{s,max} = \frac{\varepsilon_{max}}{Z} = \frac{6}{5} = 1.2 	ext{ A}$.

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