A circuit containing an 80 mH inductor and a 60 µF capacitor in series is connected to a 230 V, 50 Hz supply.

Question

A circuit containing an 80 mH inductor and a 60 µF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible. The current amplitude is:

Accepted Answer

To find the current amplitude ($i_m$), we follow these steps based on formulas in the sources: 
1. **Calculate Reactances**: The inductive reactance is $X_L = \omega L = 2\pi fL = 2 	imes 3.14 	imes 50 	imes 0.08 pprox 25.12 \, \Omega$ . The capacitive reactance is $X_C = 1/(\omega C) = 1/(2\pi fC) = 1/(2 	imes 3.14 	imes 50 	imes 60 	imes 10^{-6}) pprox 53.08 \, \Omega$ . 
2. **Calculate Impedance**: For an LC circuit with negligible resistance, impedance $Z = |X_L - X_C| = |25.12 - 53.08| = 27.96 \, \Omega$. 
3. **Calculate RMS Current**: $I_{rms} = V_{rms} / Z = 230 / 27.96 pprox 8.226$ A . 
4. **Determine Current Amplitude**: The peak current is $i_m = \sqrt{2} 	imes I_{rms} = 1.414 	imes 8.226 pprox 11.63$ A .

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