In a series `LCR` circuit with an AC source, `R = 300 Omega, C = 20 muF, L = 1.0 henry, epsilon_(rms) = 50 V` and `v = 50/(pi) Hz`. Find (a) the rms current in the circuit and (b) the rms potential differences across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as specified in the question.

### Given Data:
- Resistance, \( R = 300 \, \Omega \)
- Capacitance, \( C = 20 \, \mu F = 20 \times 10^{-6} \, F \)
- Inductance, \( L = 1.0 \, H \)
- RMS Voltage, \( \epsilon_{rms} = 50 \, V \)
- Frequency, \( f = \frac{50}{\pi} \, Hz \)

### Part (a): Finding the RMS Current in the Circuit

1. **Calculate the angular frequency (\( \omega \))**:
\[ \omega = 2 \pi f = 2 \pi \left(\frac{50}{\pi}\right) = 100 \, rad/s \]

2. **Calculate the reactance of the capacitor (\( X_C \))**:
\[ X_C = \frac{1}{\omega C} = \frac{1}{100 \times 20 \times 10^{-6}} = \frac{1}{0.002} = 500 \, \Omega \]

3. **Calculate the reactance of the inductor (\( X_L \))**:
\[ X_L = \omega L = 100 \times 1 = 100 \, \Omega \]

4. **Calculate the impedance (\( Z \)) of the circuit**:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
\[ Z = \sqrt{300^2 + (100 - 500)^2} = \sqrt{300^2 + (-400)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega \]

5. **Calculate the peak voltage (\( E_0 \))**:
\[ E_0 = \sqrt{2} \cdot \epsilon_{rms} = \sqrt{2} \cdot 50 \approx 70.71 \, V \]

6. **Calculate the RMS current (\( I_{rms} \))**:
\[ I_{rms} = \frac{E_{rms}}{Z} = \frac{50}{500} = 0.1 \, A \]

### Part (b): Finding the RMS Potential Differences

1. **Calculate the RMS potential difference across the capacitor (\( V_C \))**:
\[ V_C = I_{rms} \cdot X_C = 0.1 \cdot 500 = 50 \, V \]

2. **Calculate the RMS potential difference across the resistor (\( V_R \))**:
\[ V_R = I_{rms} \cdot R = 0.1 \cdot 300 = 30 \, V \]

3. **Calculate the RMS potential difference across the inductor (\( V_L \))**:
\[ V_L = I_{rms} \cdot X_L = 0.1 \cdot 100 = 10 \, V \]

4. **Calculate the total potential difference across all components**:
\[ V_{total} = V_R + V_C + V_L = 30 + 50 + 10 = 90 \, V \]

### Summary of Results:
- (a) The RMS current in the circuit is \( 0.1 \, A \).
- (b) The RMS potential differences are:
- Across the capacitor: \( 50 \, V \)
- Across the resistor: \( 30 \, V \)
- Across the inductor: \( 10 \, V \)
- The total potential difference across the components is \( 90 \, V \), which is greater than the RMS voltage of the source \( 50 \, V \).