An inductor `20xx10^(-3)` a capacitor `100(mu)F` and a resistor `50 Omega` are connected in series across a source of emf `V=10 sin 314 t`.
Then the energy dissipated in the circuit in 20 mim is

Options

A. 960 J
B. 900 J
C. 250J
D. 500J

Explanation

To solve the problem, we need to find the energy dissipated in the circuit over a period of 20 minutes when an inductor, capacitor, and resistor are connected in series across an alternating current source.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Inductance, \( L = 20 \times 10^{-3} \, \text{H} \)
- Capacitance, \( C = 100 \, \mu\text{F} = 100 \times 10^{-6} \, \text{F} \)
- Resistance, \( R = 50 \, \Omega \)
- Voltage source, \( V(t) = 10 \sin(314t) \)

2. **Calculate Angular Frequency (\( \omega \)):**
- From the voltage equation, we can identify \( \omega = 314 \, \text{rad/s} \).

3. **Calculate Inductive Reactance (\( X_L \)):**
\[ X_L = \omega L = 314 \times (20 \times 10^{-3}) = 6.28 \, \Omega \]

4. **Calculate Capacitive Reactance (\( X_C \)):**
\[ X_C = \frac{1}{\omega C} = \frac{1}{314 \times (100 \times 10^{-6})} \approx 31.85 \, \Omega \]

5. **Calculate Impedance (\( Z \)):**
- The total impedance in the circuit is given by:
\[ Z = \sqrt{R^2 + (X_C - X_L)^2} \]
\[ Z = \sqrt{50^2 + (31.85 - 6.28)^2} = \sqrt{2500 + (25.57)^2} \approx \sqrt{2500 + 653.0649} \approx \sqrt{3153.0649} \approx 56.14 \, \Omega \]

6. **Calculate RMS Voltage (\( V_{rms} \)):**
- The peak voltage \( V_0 = 10 \, V \), thus:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \approx 7.07 \, V \]

7. **Calculate RMS Current (\( I_{rms} \)):**
\[ I_{rms} = \frac{V_{rms}}{Z} = \frac{7.07}{56.14} \approx 0.125 \, A \]

8. **Calculate Average Power (\( P \)):**
- The average power in an AC circuit is given by:
\[ P = I_{rms}^2 R = (0.125)^2 \times 50 \approx 0.015625 \times 50 \approx 0.78125 \, W \approx 0.8 \, W \]

9. **Calculate Energy Dissipated in 20 Minutes:**
- Convert 20 minutes to seconds: \( 20 \times 60 = 1200 \, s \)
\[ \text{Energy} = P \times t = 0.8 \times 1200 \approx 960 \, J \]

### Final Answer:
The energy dissipated in the circuit in 20 minutes is **960 Joules**.

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