A coil has inductance of `(11)/(5pi)H` and is joined in series with a resistance of `220Omega`. When an alternating emf of `220V` at `50Hz` is applied to it, then the wattless component of the current in the circuit is

Options

A. `5A`
B. `0.7A`
C. `0.5A`
D. `7A`

Explanation

To solve the problem step by step, we will follow the outlined approach:

### Step 1: Identify the given values
- Inductance \( L = \frac{11}{5\pi} \, H \)
- Resistance \( R = 220 \, \Omega \)
- Alternating emf \( V = 220 \, V \)
- Frequency \( f = 50 \, Hz \)

### Step 2: Calculate the inductive reactance \( X_L \)
The inductive reactance \( X_L \) is given by the formula:
\[ X_L = 2\pi f L \]
Substituting the values:
\[ X_L = 2\pi \times 50 \times \frac{11}{5\pi} \]
Here, the \( \pi \) cancels out:
\[ X_L = 2 \times 50 \times \frac{11}{5} = 2 \times 10 \times 11 = 220 \, \Omega \]

### Step 3: Calculate the impedance \( Z \)
The impedance \( Z \) in an LR circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values of \( R \) and \( X_L \):
\[ Z = \sqrt{220^2 + 220^2} = \sqrt{2 \times 220^2} = 220\sqrt{2} \, \Omega \]

### Step 4: Calculate the phase difference \( \phi \)
The phase difference \( \phi \) is given by:
\[ \tan \phi = \frac{X_L}{R} \]
Substituting the values:
\[ \tan \phi = \frac{220}{220} = 1 \]
Thus, \( \phi = 45^\circ \).

### Step 5: Calculate the maximum current \( I_0 \)
The maximum current \( I_0 \) can be calculated using:
\[ I_0 = \frac{V}{Z} \]
Substituting the values:
\[ I_0 = \frac{220}{220\sqrt{2}} = \frac{1}{\sqrt{2}} \, A \]

### Step 6: Calculate the wattless component of the current
The wattless component of the current \( I_w \) is given by:
\[ I_w = I_0 \sin \phi \]
Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} \):
\[ I_w = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \, A = 0.5 \, A \]

### Final Answer
The wattless component of the current in the circuit is \( 0.5 \, A \).

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