In a series LCR circuit with an AC so `(E_(rms)=50V)` and `f=50//pi` Hz, R=300 ohm, C=0.02mF, L=1.0H, which of the follwing is correct

Options

A. the rms current in the circuit is 0.1 A
B. the rms potential difference acrosss the capacitor is 50V
C. the rms potential difference across the inductor is 10V
D. the rms current in the circuit is 0.14A

Answer

To solve the problem, we need to analyze the series LCR circuit with the given parameters. Let's break it down step by step.

### Step 1: Identify Given Values
- \( E_{\text{rms}} = 50 \, \text{V} \)
- Frequency \( f = \frac{50}{\pi} \, \text{Hz} \)
- Resistance \( R = 300 \, \Omega \)
- Capacitance \( C = 0.02 \, \text{mF} = 0.02 \times 10^{-3} \, \text{F} = 2 \times 10^{-5} \, \text{F} \)
- Inductance \( L = 1.0 \, \text{H} \)

### Step 2: Calculate Angular Frequency
The angular frequency \( \omega \) is given by:
\[ \omega = 2\pi f \]
Substituting the value of \( f \):
\[ \omega = 2\pi \left(\frac{50}{\pi}\right) = 100 \, \text{rad/s} \]

### Step 3: Calculate Inductive Reactance \( X_L \)
The inductive reactance \( X_L \) is calculated using the formula:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 100 \times 1.0 = 100 \, \Omega \]

### Step 4: Calculate Capacitive Reactance \( X_C \)
The capacitive reactance \( X_C \) is calculated using the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{100 \times 2 \times 10^{-5}} = \frac{1}{0.002} = 500 \, \Omega \]

### Step 5: Calculate Impedance \( Z \)
The total impedance \( Z \) in the circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values:
\[ Z = \sqrt{300^2 + (100 - 500)^2} = \sqrt{300^2 + (-400)^2} \]
Calculating:
\[ Z = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega \]

### Step 6: Calculate RMS Current \( I_{\text{rms}} \)
The RMS current can be calculated using Ohm's law:
\[ I_{\text{rms}} = \frac{E_{\text{rms}}}{Z} \]
Substituting the values:
\[ I_{\text{rms}} = \frac{50}{500} = 0.1 \, \text{A} \]

### Step 7: Calculate Potential Difference Across Capacitor \( V_C \)
The potential difference across the capacitor can be calculated as:
\[ V_C = I_{\text{rms}} \times X_C \]
Substituting the values:
\[ V_C = 0.1 \times 500 = 50 \, \text{V} \]

### Step 8: Calculate Potential Difference Across Inductor \( V_L \)
The potential difference across the inductor can be calculated as:
\[ V_L = I_{\text{rms}} \times X_L \]
Substituting the values:
\[ V_L = 0.1 \times 100 = 10 \, \text{V} \]

### Summary of Results
- \( I_{\text{rms}} = 0.1 \, \text{A} \)
- \( V_C = 50 \, \text{V} \)
- \( V_L = 10 \, \text{V} \)

### Conclusion
Based on the calculations:
- Option A is correct (RMS current is 0.1 A).
- Option B is correct (Potential difference across capacitor is 50 V).
- Option C is correct (Potential difference across inductor is 10 V).
- Option D is incorrect (RMS current is not 0.14 A).