To solve the problem step by step, we will follow the outlined approach in the video transcript.
### 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, \( v = \frac{50}{\pi} \, Hz \)
### Step 1: Calculate Angular Frequency
The angular frequency \( \omega \) is given by:
\[
\omega = 2 \pi v = 2 \pi \left(\frac{50}{\pi}\right) = 100 \, rad/s
\]
### Step 2: Calculate Reactance of Capacitor \( X_C \)
The capacitive reactance \( X_C \) is given by:
\[
X_C = \frac{1}{\omega C}
\]
Substituting the values:
\[
X_C = \frac{1}{100 \times 20 \times 10^{-6}} = \frac{1}{0.002} = 500 \, \Omega
\]
### Step 3: Calculate Reactance of Inductor \( X_L \)
The inductive reactance \( X_L \) is given by:
\[
X_L = \omega L
\]
Substituting the values:
\[
X_L = 100 \times 1 = 100 \, \Omega
\]
### Step 4: Calculate Total Impedance \( Z \)
The total impedance \( Z \) in a series LCR circuit is given by:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
Calculating \( X_L - X_C \):
\[
X_L - X_C = 100 - 500 = -400 \, \Omega
\]
Now substituting into the impedance formula:
\[
Z = \sqrt{300^2 + (-400)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega
\]
### Step 5: Calculate RMS Current \( I_{rms} \)
The RMS current in the circuit is given by:
\[
I_{rms} = \frac{\epsilon_{rms}}{Z}
\]
Substituting the values:
\[
I_{rms} = \frac{50}{500} = 0.1 \, A
\]
### Step 6: Calculate RMS Voltage Across Each Component
1. **Voltage across Capacitor \( V_C \)**:
\[
V_C = I_{rms} \times X_C = 0.1 \times 500 = 50 \, V
\]
2. **Voltage across Resistor \( V_R \)**:
\[
V_R = I_{rms} \times R = 0.1 \times 300 = 30 \, V
\]
3. **Voltage across Inductor \( V_L \)**:
\[
V_L = I_{rms} \times X_L = 0.1 \times 100 = 10 \, V
\]
### Step 7: Total Voltage Across All Components
The total voltage across all components is:
\[
V_{total} = V_C + V_R + V_L = 50 + 30 + 10 = 90 \, V
\]
### Conclusion
- (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 sum of the RMS potential differences \( 90 \, V \) is greater than the RMS voltage of the source \( 50 \, V \).
