Alternating e.m.f. of `E = 220 sin 100 pi t` is applied to a circuit containing an inductance of `(1//pi)` henry. Write equation for instantaneous current through the circuit. What will be the reading of a.c. galvanometer connected in the circuit ?

Explanation

To solve the problem step by step, we will derive the equation for the instantaneous current through the circuit and then calculate the reading of the AC galvanometer.

### Step 1: Identify the given parameters
The alternating e.m.f. is given by:
\[ E(t) = 220 \sin(100 \pi t) \]
From this, we can identify:
- \( E_0 = 220 \, \text{V} \) (peak voltage)
- \( \omega = 100 \pi \, \text{rad/s} \) (angular frequency)
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