A current of `4A` flows in a coil when connected to a `12V DC` source. If the same coil is connected to a `12V, 50 rad//s AC` source, a current of `2.4A` flows in the circuit. Determine the inductance of the coil. Also, find the power developed in the circuit if a`2500muF` capacitor is connected in series with the coil.

Options

A. `(pi/sqrt3)H`
B. `(sqrt3/pi)H`
C. `(2/pi)H`
D. none of these

Explanation

To solve the problem, we need to find the inductance of the coil and the power developed in the circuit when a capacitor is connected in series with the coil.

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

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

### Step 2: Calculate the Inductive Reactance and Impedance
When the coil is connected to a 12V, 50 rad/s AC source, the current flowing is 2.4A. We can use the formula for the RMS current in an AC circuit:

\[ I_{rms} = \frac{V_{rms}}{Z} \]

Where \(Z\) is the impedance of the circuit. The impedance for a circuit with resistance and inductance is given by:

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

Where \(X_L\) (inductive reactance) is given by:

\[ X_L = \omega L \]

### Step 3: Rearranging the Formula
From the equation \(I_{rms} = \frac{V_{rms}}{Z}\), we can rearrange to find \(Z\):

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

### Step 4: Set Up the Equation for Impedance
Now, we can set up the equation for impedance:

\[ 5 = \sqrt{3^2 + (50L)^2} \]

### Step 5: Square Both Sides
Squaring both sides gives:

\[ 25 = 9 + (50L)^2 \]

### Step 6: Solve for \(L\)
Rearranging gives:

\[ (50L)^2 = 25 - 9 = 16 \]

\[ 50L = 4 \implies L = \frac{4}{50} = 0.08 \, H \]

### Step 7: Calculate the Power Developed in the Circuit
Now we need to find the power developed in the circuit when a 2500 µF capacitor is connected in series with the coil. The power \(P\) in an AC circuit is given by:

\[ P = V_{rms} \cdot I_{rms} \cdot \cos \phi \]

Where \(\cos \phi\) is the power factor. The power factor can be calculated as:

\[ \cos \phi = \frac{R}{Z} \]

### Step 8: Calculate the Power Factor
Substituting the values we have:

\[ \cos \phi = \frac{3}{5} = 0.6 \]

### Step 9: Calculate the Power
Now substituting into the power formula:

\[ P = 12 \cdot 2.4 \cdot 0.6 = 17.28 \, W \]

### Final Answers
- The inductance of the coil \(L = 0.08 \, H\)
- The power developed in the circuit \(P = 17.28 \, W\)

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