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An inductor of 20mH, a capacitor 100 µF and a resistor 50 Ω are connected in a series across a source of e.m.f. V = 10 sin (314t). Find the energy dissipated in the circuit in 20 minutes. If resistance is removed from the circuit and the value of inductance is doubled, then find the variation of current with time in the new circuit.

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For the LCR circuit, the energy dissipated over a long time is \(U = (V_{rms}I_{rms}cos\phi)t\). When resistance is removed, the circuit becomes LC circuit, the impedance and hence current changes.

The circuit is as shown in figure. One time cycle \(T = \frac{2\pi}\omega = \frac{2\pi}{314} = 0.02s\). So, we have to calculate the average energy at time t>>T.

Energy dissipated in time t

When resistance is removed, and inductance is doubled, then cosφ = 0 ⇒ φ = π/2

Value of impedance is

And the current in the circuit is found to be

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