An a.c. voltage given by relation, \(V = 300\sin(100t)\text{ volt}\) is connected with resistor having resistance \(60\ \Omega\) and inductor of inductance \(L\). If peak current in the circuit is \(3\text{ A}\), the value of \(L\) will be (where \(t\) denotes the time in s)
Solution:
Peak voltage \(V_0 = 300\text{ V}\), peak current \(I_0 = 3\text{ A}\), so impedance \(Z = V_0/I_0 = 100\ \Omega\). Using \(Z = \sqrt{R^2 + (\omega L)^2}\), we have \(100 = \sqrt{60^2 + (100L)^2}\), which gives \(100L = 80 \implies L = 0.8\text{ H}\).
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