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Alternating e.m.f. of E = 220 sin 100 pi...

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 ?

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AI Generated Solution

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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Knowledge Check

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