A circuit operating at `360/(2pi) Hz` contains a `1muF` capacitor and a `20 Omega`. resistor. How large an inductor must be added in series to make the phase angle for the circuit zero? Calculate the current in the circuit if the applied voltage is `120 V`.

Explanation

To solve the problem step by step, we need to determine the inductance required to make the phase angle zero in the given circuit and then calculate the current flowing through the circuit when a voltage of 120 V is applied.

### Step 1: Identify the Given Values
- Frequency \( f = \frac{360}{2\pi} \) Hz
- Capacitance \( C = 1 \mu F = 1 \times 10^{-6} F \)
- Resistance \( R = 20 \Omega \)
- Voltage \( V = 120 V \)

### Step 2: Calculate Angular Frequency \( \omega \)
The angular frequency \( \omega \) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of \( f \):
\[ \omega = 2\pi \left(\frac{360}{2\pi}\right) = 360 \, \text{rad/s} \]

### Step 3: Set Up the Condition for Zero Phase Angle
For the phase angle \( \phi \) to be zero, the circuit must be at resonance, which means the inductive reactance \( X_L \) must equal the capacitive reactance \( X_C \):
\[ X_L = X_C \]
Where:
\[ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} \]

### Step 4: Equate Inductive and Capacitive Reactance
Setting \( X_L \) equal to \( X_C \):
\[ \omega L = \frac{1}{\omega C} \]
From this, we can solve for \( L \):
\[ L = \frac{1}{\omega^2 C} \]

### Step 5: Substitute the Values to Find \( L \)
Substituting \( \omega = 360 \, \text{rad/s} \) and \( C = 1 \times 10^{-6} F \):
\[ L = \frac{1}{(360)^2 \times (1 \times 10^{-6})} \]
Calculating \( (360)^2 \):
\[ (360)^2 = 129600 \]
Now substituting back:
\[ L = \frac{1}{129600 \times 10^{-6}} = \frac{1}{0.1296} \approx 7.72 \, H \]

### Step 6: Calculate the Current in the Circuit
At resonance, the impedance \( Z \) is equal to the resistance \( R \):
\[ Z = R = 20 \, \Omega \]
Using Ohm's Law to calculate the RMS current \( I \):
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Substituting the values:
\[ I_{rms} = \frac{120}{20} = 6 \, A \]

### Final Answers
- Inductance \( L \approx 7.72 \, H \)
- RMS Current \( I_{rms} = 6 \, A \)