Q.An alternating emf $e = 220\sin(100\pi t)$ is applied to a circuit containing an inductance of $\frac{1}{\pi}$ henry. Write an equation for the instantaneous current through the circuit. What will be the reading of the AC galvanometer connected in the circuit?

[!TLDR] $X_L=\omega L=100\pi\times(1/\pi)=100\,\Omega$; $i_0=e_0/X_L=220/100=2.2$ A; current lags by $\pi/2$: $i=2.2\sin(100\pi t-\pi/2)$. Galvanometer (rms) reading $=i_0/\sqrt2\approx1 …

Given $e=220\sin(100\pi t)$, so $e_0=220$ V and $\omega=100\pi$ rad/s. With $L=\dfrac{1}{\pi}$ H, the inductive reactance is $X_L=\omega L=100\pi\times\dfrac{1}{\pi}=100\,\Omega$.\n\nThe peak current is $i_0=\dfrac{e_0}{X_L}=\dfrac{220}{100}=2.2$ A. Since current lags emf by $\pi/2$ in a pure inductor, the instantaneous current is $i=i_0\sin\left(\omega t-\dfrac{\pi}{2}\right)=2.2\sin\left(100\pi t-\dfrac{\pi}{2}\right)$ A.\n\nAn AC g …

Compute $X_L=\omega L$ from the given emf's angular frequency, find $i_0=e_0/X_L$, apply the pure-inductor $\pi/2$ lag to …

Writing the current with a $+\pi/2$ (lead) instead of $-\pi/2$ (lag), which is the inductor's signature phase relati …