A coil has `L=0.04H` and `R=12Omega`. When it is connected to `220V, 50Hz` supply the current flowing through the coil, in ampere is

Options

A. 12.7 A
B. 14.7 A
C. 11.7 A
D. 10.7 A

Answer

To find the current flowing through the coil when it is connected to a 220V, 50Hz supply, we will follow these steps:

### Step 1: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
where \( f \) is the frequency in hertz.

Given:
- \( f = 50 \, \text{Hz} \)

Calculating ω:
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

### Step 2: Calculate the inductive reactance (XL)
The inductive reactance (XL) is given by the formula:
\[ X_L = \omega L \]
where \( L \) is the inductance in henries.

Given:
- \( L = 0.04 \, \text{H} \)

Calculating \( X_L \):
\[ X_L = 100\pi \times 0.04 = 4\pi \, \Omega \]

### Step 3: Calculate the impedance (Z)
The impedance (Z) in an R-L circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]

Given:
- \( R = 12 \, \Omega \)

Calculating \( Z \):
\[ Z = \sqrt{12^2 + (4\pi)^2} \]
Calculating \( 12^2 \):
\[ 12^2 = 144 \]
Calculating \( (4\pi)^2 \):
\[ (4\pi)^2 = 16\pi^2 \approx 16 \times 10 \approx 160 \quad (\text{using } \pi^2 \approx 10) \]
Now substituting back:
\[ Z = \sqrt{144 + 160} = \sqrt{304} \approx 17.43 \, \Omega \]

### Step 4: Calculate the current (I)
The current (I) can be calculated using Ohm's law:
\[ I = \frac{V}{Z} \]

Given:
- \( V = 220 \, \text{V} \)

Calculating \( I \):
\[ I = \frac{220}{17.43} \approx 12.6 \, \text{A} \]

### Final Answer:
The current flowing through the coil is approximately \( 12.6 \, \text{A} \).

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