When an inductor coil is connected to an ideal battery of emf 10V, a constant current 2.5 A flows. When the same inductor coil is connected to an ac source of 10V and 50 HZ then the current is 2A. Find out the inductance of the coil.

Explanation

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

### Step 1: Calculate the Resistance of the Coil
When the inductor coil is connected to a DC source (battery), we can use Ohm's law to find the resistance (R) of the coil.

Using Ohm's law:
\[ V = I \times R \]

Where:
- \( V = 10 \, \text{V} \) (emf of the battery)
- \( I = 2.5 \, \text{A} \) (current flowing through the coil)

Rearranging the formula to find R:
\[ R = \frac{V}{I} = \frac{10 \, \text{V}}{2.5 \, \text{A}} = 4 \, \Omega \]

### Step 2: Calculate the Impedance of the Coil
When the same coil is connected to an AC source, we can calculate the impedance (Z) using the formula:
\[ Z = \frac{V}{I} \]

Where:
- \( V = 10 \, \text{V} \) (voltage of the AC source)
- \( I = 2 \, \text{A} \) (current flowing through the coil)

Calculating Z:
\[ Z = \frac{10 \, \text{V}}{2 \, \text{A}} = 5 \, \Omega \]

### Step 3: Relate Impedance, Resistance, and Inductive Reactance
The impedance of the coil in an AC circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]

Where:
- \( X_L \) is the inductive reactance.

Squaring both sides gives:
\[ Z^2 = R^2 + X_L^2 \]

Substituting the values we found:
\[ 5^2 = 4^2 + X_L^2 \]
\[ 25 = 16 + X_L^2 \]

### Step 4: Solve for Inductive Reactance
Rearranging gives:
\[ X_L^2 = 25 - 16 = 9 \]
Taking the square root:
\[ X_L = 3 \, \Omega \]

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

Where:
\[ \omega = 2 \pi f \]
And given \( f = 50 \, \text{Hz} \):
\[ \omega = 2 \pi \times 50 = 100 \pi \]

Now substituting \( X_L \) into the equation:
\[ 3 = (100 \pi) L \]

Rearranging to find \( L \):
\[ L = \frac{3}{100 \pi} \]

### Final Answer
The inductance of the coil is:
\[ L = \frac{3}{100 \pi} \, \text{H} \]

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