Find the energy dissipated in a time interval of 200 s when series AC circuit of 20 mH inductance 100 `muF` capacitance and 50 `Omega` resistance is connected to an AC source 12V (rms ) and 50 Hz.

Explanation

To find the energy dissipated in the given AC circuit, we will follow these steps:

### Step 1: Convert given values to standard units
- Inductance \( L = 20 \, \text{mH} = 20 \times 10^{-3} \, \text{H} \)
- Capacitance \( C = 100 \, \mu\text{F} = 100 \times 10^{-6} \, \text{F} \)
- Resistance \( R = 50 \, \Omega \)
- RMS Voltage \( V_{\text{RMS}} = 12 \, \text{V} \)
- Frequency \( f = 50 \, \text{Hz} \)
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