A coil of resistance 52 and inductance 120mH in series with a 100 µF capacitor is connected to a 300 V, 50 Hz supply. Calculate (a) the current flowing, (b) the phase difference between the supply voltage and current. (c) the voltage across the coil and (d) the voltage across the capacitor.

To solve this problem, we'll apply the concepts of AC circuits, particularly those involving resistance, inductance, and capacitance.

### Given Data:
- Resistance, \( R = 52 \, \Omega \)
- Inductance, \( L = 120 \, \text{mH} = 0.12 \, \text{H} \)
- Capacitance, \( C = 100 \, \mu\text{F} = 100 \times 10^{-6} \, \text{F} \)
- Supply voltage, \( V_s = 300 \, \text{V} \)
- Frequency, \( f = 50 \, \text{Hz} \)

### Formulas:

1. **Inductive Reactance**: 
   \[
   X_L = 2\pi f L
   \]
2. **Capacitive Reactance**: 
   \[
   X_C = \frac{1}{2\pi f C}
   \]
3. **Impedance**: 
   \[
   Z = \sqrt{R^2 + (X_L - X_C)^2}
   \]
4. **Current**: 
   \[
   I = \frac{V_s}{Z}
   \]
5. **Phase Angle**: 
   \[
   \phi = \arctan\left(\frac{X_L - X_C}{R}\right)
   \]
6. **Voltage across the Coil** (Resistor + Inductor): 
   \[
   V_{\text{coil}} = I \times Z_{\text{coil}}
   \]
   where \( Z_{\text{coil}} = \sqrt{R^2 + X_L^2} \)
7. **Voltage across the Capacitor**: 
   \[
   V_C = I \times X_C
   \]

### Step-by-Step Solution:

1. **Calculate Inductive Reactance**:
   \[
   X_L = 2 \pi \times 50 \times 0.12 = 37.7 \, \Omega
   \]

2. **Calculate Capacitive Reactance**:
   \[
   X_C = \frac{1}{2 \pi \times 50 \times 100 \times 10^{-6}} = 31.8 \, \Omega
   \]

3. **Calculate Impedance**:
   \[
   Z = \sqrt{52^2 + (37.7 - 31.8)^2} = \sqrt{52^2 + 5.9^2} = \sqrt{2704 + 34.81} \approx 52.33 \, \Omega
   \]

4. **Calculate the Current**:
   \[
   I = \frac{300}{52.33} \approx 5.73 \, \text{A}
   \]

5. **Calculate the Phase Angle**:
   \[
   \phi = \arctan\left(\frac{37.7 - 31.8}{52}\right) = \arctan\left(\frac{5.9}{52}\right) \approx \arctan(0.113) \approx 6.45^\circ
   \]
   The current lags the voltage by 6.45°.

6. **Calculate Voltage across the Coil**:
   \[
   Z_{\text{coil}} = \sqrt{52^2 + 37.7^2} \approx \sqrt{2704 + 1420.29} \approx \sqrt{4124.29} \approx 64.2 \, \Omega
   \]
   \[
   V_{\text{coil}} = 5.73 \times 64.2 \approx 367.9 \, \text{V}
   \]

7. **Calculate Voltage across the Capacitor**:
   \[
   V_C = 5.73 \times 31.8 \approx 182.2 \, \text{V}
   \]

### Final Answers:
- **(a) The current flowing**: \( \approx 5.73 \, \text{A} \)
- **(b) The phase difference**: \( \approx 6.45^\circ \) (current lags the voltage)
- **(c) Voltage across the coil**: \( \approx 367.9 \, \text{V} \)
- **(d) Voltage across the capacitor**: \( \approx 182.2 \, \text{V} \)

Would you like more details on any of these steps or do you have any further questions?

Here are 5 related questions for further exploration:
1. What happens to the phase angle if the inductance increases?
2. How would the current change if the resistance of the coil was doubled?
3. What would the resonance frequency of this circuit be?
4. How does the voltage across the capacitor change if the frequency of the supply is increased?
5. What effect does reducing the capacitance to 50 µF have on the impedance of the circuit?

**Tip:** Always check the phase relationship in AC circuits to understand the behavior of the voltage and current over time.

Learn how to analyze an AC circuit with a coil of resistance 52Ω and inductance 120mH in series with a 100 µF capacitor connected to a 300 V, 50 Hz supply. Calculate current, phase difference, voltage across the coil, and voltage across the capacitor. Detailed step-by-step solution provided.

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