An AC circuit having supply voltage E = 10 si | vedclass

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(C) The supply voltage is $E = 10 \sin \omega t$. The circuit is an $RL$ series circuit.The impedance of the circuit is $Z = \sqrt{R^2 + X_L^2} = \sqr

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$4.8$

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(C) The supply voltage is $E = 10 \sin \omega t$. The circuit is an $RL$ series circuit.The impedance of the circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{3^2 + 4^2} = 5 \, \Omega$.The current in the circuit is given by $I = I_0 \sin(\omega t - \phi)$,where $I_0 = E_0 / Z = 10 / 5 = 2 \, 	ext{A}$ and $	an \phi = X_L / R = 4/3$,so $\phi = 	an^{-1}(4/3)$.The voltage across the inductor is $V_L = I X_L = (I_0 \sin(\omega t - \phi + \pi/2)) X_L = I_0 X_L \sin(\omega t - \phi + \pi/2)$.At $t = \pi / \omega$,the voltage across the inductor is $V_L = I_0 X_L \sin(\omega \cdot \frac{\pi}{\omega} - \phi + \frac{\pi}{2}) = I_0 X_L \sin(\pi - \phi + \frac{\pi}{2}) = I_0 X_L \sin(\frac{3\pi}{2} - \phi) = -I_0 X_L \cos \phi$.Since $\cos \phi = R / Z = 3 / 5 = 0.6$,we have $V_L = -(2)(4)(0.6) = -4.8 \, 	ext{V}$.The magnitude of the voltage across the inductor is $|V_L| = 4.8 \, 	ext{V}$.

An $AC$ circuit having supply voltage $E = 10 \sin \omega t$ consists of a resistor of resistance $R = 3 \Omega$ and an inductor of reactance $X_L = 4 \Omega$ as shown in the figure. The voltage across the inductor at $t = \pi / \omega$ is: (in $\text{ V}$)

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