A 4 `muF ` capacitor and a 500 `Omega` resistor are connected in series a power supply 20 V , 30 Hz . Calculate the phase angle between voltage and current .

Answer

To solve the problem of finding the phase angle between voltage and current in a series RC circuit with a capacitor of 4 µF and a resistor of 500 Ω connected to a 20 V, 30 Hz power supply, we will follow these steps:

### Step 1: Identify the given values
- Capacitance, \( C = 4 \, \mu F = 4 \times 10^{-6} \, F \)
- Resistance, \( R = 500 \, \Omega \)
- Frequency, \( f = 30 \, Hz \)

### Step 2: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is given by the formula:
\[ \omega = 2 \pi f \]
Substituting the value of \( f \):
\[ \omega = 2 \pi \times 30 \approx 188.4 \, rad/s \]

### Step 3: Calculate the capacitive reactance \( X_c \)
The capacitive reactance \( X_c \) is given by the formula:
\[ X_c = \frac{1}{\omega C} \]
Substituting the values of \( \omega \) and \( C \):
\[ X_c = \frac{1}{188.4 \times 4 \times 10^{-6}} \approx \frac{1}{7.536 \times 10^{-4}} \approx 1320.6 \, \Omega \]

### Step 4: Calculate the impedance \( Z \)
The impedance \( Z \) in an RC circuit is given by:
\[ Z = \sqrt{R^2 + X_c^2} \]
Substituting the values of \( R \) and \( X_c \):
\[ Z = \sqrt{500^2 + (1320.6)^2} \approx \sqrt{250000 + 1740000} \approx \sqrt{1990000} \approx 1414.2 \, \Omega \]

### Step 5: Calculate the phase angle \( \phi \)
The phase angle \( \phi \) between the voltage and current is given by:
\[ \phi = \tan^{-1}\left(\frac{X_c}{R}\right) \]
Substituting the values of \( X_c \) and \( R \):
\[ \phi = \tan^{-1}\left(\frac{1320.6}{500}\right) \approx \tan^{-1}(2.6412) \approx 69.4^\circ \]

### Final Result
The phase angle \( \phi \) between the voltage and current is approximately \( 69.4^\circ \).

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