A student connects a long air cored - coil of manganin wire to a 100 V D.C. supply and records a current of 25 amp. When the same coil is connected across 100 V. 50 Hz A.C. the current reduces to 20 A , the reactance of the coil is :-

Options

A. `4 Omega`
B. `3 Omega`
C. `5 Omega`
D. None

Answer

To solve the problem, we need to find the reactance (X_L) of the coil when it is connected to an AC source. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the Resistance (R)
When the coil is connected to a 100 V DC supply, the current is 25 A. Using Ohm's law (V = IR), we can find the resistance of the coil.

\[ R = \frac{V}{I} = \frac{100 \, \text{V}}{25 \, \text{A}} = 4 \, \Omega \]

### Step 2: Set Up the AC Circuit
When the same coil is connected to a 100 V AC supply at 50 Hz, the current reduces to 20 A. In this case, the circuit consists of the resistance (R) and the inductive reactance (X_L).

### Step 3: Calculate the Impedance (Z)
The total impedance (Z) in the circuit can be expressed as:

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

### Step 4: Apply Ohm's Law for AC Circuit
Using Ohm's law for the AC circuit, we can express the current (I) in terms of voltage (V) and impedance (Z):

\[ I = \frac{V}{Z} \]

Substituting the known values:

\[ 20 \, \text{A} = \frac{100 \, \text{V}}{Z} \]

From this, we can find Z:

\[ Z = \frac{100 \, \text{V}}{20 \, \text{A}} = 5 \, \Omega \]

### Step 5: Substitute Z into the Impedance Equation
Now we can substitute the value of Z back into the impedance equation:

\[ 5 = \sqrt{4^2 + X_L^2} \]

### Step 6: Solve for X_L
Squaring both sides gives:

\[ 25 = 16 + X_L^2 \]

Rearranging gives:

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

Taking the square root:

\[ X_L = 3 \, \Omega \]

### Final Answer
The reactance of the coil is **3 Ω**.

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