A coil has reactance of `100Omega` when frequency is 50Hz. If the frequency becomes 150Hz, then the reactance will be –

Options

A. `100Omega`
B. `300Omega`
C. `450Omega`
D. `600Omega`

Answer

To solve the problem, we need to find the reactance of the coil when the frequency changes from 50 Hz to 150 Hz. The reactance \(X_L\) of an inductor is given by the formula:

\[ X_L = 2 \pi f L \]

where:
- \(X_L\) is the inductive reactance,
- \(f\) is the frequency,
- \(L\) is the inductance of the coil.

### Step 1: Identify the given values
We know:
- Initial reactance \(X_L = 100 \, \Omega\) at frequency \(f = 50 \, \text{Hz}\).

### Step 2: Write the reactance formula for the initial condition
Using the formula for reactance, we can express it for the initial condition:
\[ 100 = 2 \pi (50) L \]

### Step 3: Solve for inductance \(L\)
Rearranging the equation to find \(L\):
\[ L = \frac{100}{2 \pi (50)} = \frac{100}{100\pi} = \frac{1}{\pi} \, \text{H} \]

### Step 4: Write the reactance formula for the new frequency
Now, we need to find the reactance at the new frequency \(f = 150 \, \text{Hz}\):
\[ X_L' = 2 \pi (150) L \]

### Step 5: Substitute \(L\) into the new reactance formula
Substituting \(L = \frac{1}{\pi}\):
\[ X_L' = 2 \pi (150) \left(\frac{1}{\pi}\right) = 2 \times 150 = 300 \, \Omega \]

### Step 6: Conclusion
The reactance of the coil when the frequency is 150 Hz is:
\[ X_L' = 300 \, \Omega \]

### Final Answer
The reactance will be \(300 \, \Omega\).

---