A coil has a inductance of `0.7pi` H and is joined in series with a resistance of `220 Omega`. When an alternating emf of 220V at 50 cps is applied to it, then the watt-less component of the current in the circuit is `(take 0.7pi = 2.2)`

Options

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

Explanation

To solve the problem step by step, we need to find the wattless component of the current in a circuit consisting of a coil with inductance and resistance in series when an alternating emf is applied.

### Given Data:
- Inductance, \( L = 0.7\pi \, \text{H} \) (which is approximately \( 2.2 \, \text{H} \))
- Resistance, \( R = 220 \, \Omega \)
- Alternating emf, \( V = 220 \, \text{V} \)
- Frequency, \( f = 50 \, \text{Hz} \)

### Step 1: Calculate the inductive reactance (\( X_L \))
The inductive reactance is given by the formula:
\[ X_L = L \cdot \omega \]
where \( \omega = 2\pi f \).

Calculating \( \omega \):
\[ \omega = 2\pi \cdot 50 = 100\pi \, \text{rad/s} \]

Now, substituting the values:
\[ X_L = 2.2 \cdot 100\pi = 220 \, \Omega \]

### Step 2: Calculate the impedance (\( Z \))
The impedance in a series circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]

Substituting the values:
\[ Z = \sqrt{220^2 + 220^2} = \sqrt{2 \cdot 220^2} = 220\sqrt{2} \, \Omega \]

### Step 3: Calculate the RMS current (\( I_{\text{rms}} \))
The RMS current can be calculated using Ohm's law for AC circuits:
\[ I_{\text{rms}} = \frac{V}{Z} \]

Substituting the values:
\[ I_{\text{rms}} = \frac{220}{220\sqrt{2}} = \frac{1}{\sqrt{2}} \, \text{A} \]

### Step 4: Calculate the phase angle (\( \phi \))
The phase angle \( \phi \) can be calculated using:
\[ \tan \phi = \frac{X_L}{R} \]

Substituting the values:
\[ \tan \phi = \frac{220}{220} = 1 \]
Thus,
\[ \phi = \tan^{-1}(1) = 45^\circ \]

### Step 5: Calculate the wattless component of the current (\( I' \))
The wattless component of the current is given by:
\[ I' = I_{\text{rms}} \cdot \sin \phi \]

Substituting the values:
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \]
Thus,
\[ I' = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \, \text{A} = 0.5 \, \text{A} \]

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

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