A 0.21-H inductor and a `88-Omega` resistor are connected in series to a 220-V, 50-Hz AC source. The current in the circuit and the phase angle between the current and the source voltage are respectively. `(Use pi= 22//7)`

Options

A. `2A, tan^(-1)3//4`
B. `14.4A, tan^(-1)7//8`
C. `14.4A, tan^(-1)8//7`
D. `3.28 A, tan^(-1)2//11`

Explanation

To solve the problem, we need to find the current in the circuit and the phase angle between the current and the source voltage when a 0.21-H inductor and an 88-ohm resistor are connected in series to a 220-V, 50-Hz AC source.

### Step 1: Calculate the inductive reactance (X_L)
The inductive reactance \(X_L\) is given by the formula:
\[ X_L = 2 \pi f L \]
Where:
- \(f = 50 \, \text{Hz}\) (frequency)
- \(L = 0.21 \, \text{H}\) (inductance)

Substituting the values:
\[ X_L = 2 \times \frac{22}{7} \times 50 \times 0.21 \]
Calculating this:
\[ X_L = 2 \times \frac{22}{7} \times 50 \times 0.21 = \frac{2200 \times 0.21}{7} = \frac{462}{7} \approx 66 \, \Omega \]

### Step 2: Calculate the impedance (Z)
The total impedance \(Z\) in a series circuit with resistance \(R\) and inductive reactance \(X_L\) is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Where:
- \(R = 88 \, \Omega\)
- \(X_L \approx 66 \, \Omega\)

Substituting the values:
\[ Z = \sqrt{88^2 + 66^2} \]
Calculating this:
\[ Z = \sqrt{7744 + 4356} = \sqrt{12100} = 110 \, \Omega \]

### Step 3: Calculate the current (I)
The current \(I\) in the circuit can be calculated using Ohm's law:
\[ I = \frac{V}{Z} \]
Where:
- \(V = 220 \, \text{V}\)

Substituting the values:
\[ I = \frac{220}{110} = 2 \, \text{A} \]

### Step 4: Calculate the phase angle (φ)
The phase angle \(φ\) between the current and the voltage is given by:
\[ \phi = \tan^{-1}\left(\frac{X_L}{R}\right) \]
Substituting the values:
\[ \phi = \tan^{-1}\left(\frac{66}{88}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]

### Final Results
The current in the circuit is \(2 \, \text{A}\) and the phase angle \(φ\) is \(\tan^{-1}\left(\frac{3}{4}\right)\).