A current of 4 A flows in a coil when connected to a 12 V dc source. If the same coil is connected to a `12 V, 50 rad s^(-1)` ac source, a current of 2.4 A flows of the coil in the circuit.
The inductance of the coil is

Options

A. 0.02H
B. 0.04H
C. 0.08H
D. 1.0H

Explanation

To find the inductance of the coil, we will follow these steps:

### Step 1: Calculate the Resistance of the Coil
When the coil is connected to a DC source, the current flowing through it is given as 4 A with a voltage of 12 V. We can use Ohm's law to find the resistance (R) of the coil.

\[ R = \frac{V}{I} = \frac{12 \, \text{V}}{4 \, \text{A}} = 3 \, \Omega \]

### Step 2: Calculate the Impedance in AC Circuit
When the coil is connected to the AC source, the RMS voltage (V_rms) is 12 V and the current (I_rms) is 2.4 A. We can calculate the impedance (Z) using Ohm's law.

\[ Z = \frac{V_{rms}}{I_{rms}} = \frac{12 \, \text{V}}{2.4 \, \text{A}} = 5 \, \Omega \]

### Step 3: Relate Impedance to Resistance and Reactance
In an LR circuit, the impedance can be expressed as:

\[ Z = \sqrt{R^2 + X_L^2} \]

Where \(X_L\) is the inductive reactance. We can square both sides to eliminate the square root:

\[ Z^2 = R^2 + X_L^2 \]

Substituting the values we have:

\[ 5^2 = 3^2 + X_L^2 \]

\[ 25 = 9 + X_L^2 \]

### Step 4: Solve for Inductive Reactance
Rearranging the equation gives us:

\[ X_L^2 = 25 - 9 = 16 \]

Taking the square root:

\[ X_L = 4 \, \Omega \]

### Step 5: Calculate Inductance
The inductive reactance \(X_L\) is related to the inductance (L) and the angular frequency (\(\omega\)) by the formula:

\[ X_L = \omega L \]

We can rearrange this to find L:

\[ L = \frac{X_L}{\omega} = \frac{4 \, \Omega}{50 \, \text{rad/s}} = 0.08 \, \text{H} \]

### Final Answer
The inductance of the coil is:

\[ \boxed{0.08 \, \text{H}} \]

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