A series LCR circuit driven by 300V at a frequency of 50 Hz contains a resistance `R=3kOmega` an inductor of inductive reactance `X_L = 250 pi Omega` and an unknown capacitor. The value of capacitance to maximize the average power should be (Take `pi^2 = 10` )

Options

A. `400 muF`
B. `4 muF`
C. `40 muF`
D. `25 muF`

Explanation

To find the value of capacitance that maximizes the average power in a series LCR circuit, we need to follow these steps:

### Step 1: Understand Resonance Condition
In a series LCR circuit, the average power is maximized at resonance. At resonance, the inductive reactance \(X_L\) is equal to the capacitive reactance \(X_C\).

### Step 2: Write the Expression for Reactances
The inductive reactance is given as:
\[ X_L = 250\pi \, \Omega \]
The capacitive reactance is given by:
\[ X_C = \frac{1}{\omega C} \]
where \(\omega = 2\pi f\) and \(f\) is the frequency.

### Step 3: Calculate \(\omega\)
Given that the frequency \(f = 50 \, \text{Hz}\):
\[ \omega = 2\pi f = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

### Step 4: Set \(X_L\) Equal to \(X_C\)
At resonance, set \(X_L = X_C\):
\[ 250\pi = \frac{1}{\omega C} \]
Substituting \(\omega\):
\[ 250\pi = \frac{1}{100\pi C} \]

### Step 5: Solve for Capacitance \(C\)
Rearranging the equation to solve for \(C\):
\[ C = \frac{1}{250\pi \times 100\pi} \]
\[ C = \frac{1}{25000\pi^2} \]

### Step 6: Substitute \(\pi^2\)
Given that \(\pi^2 = 10\):
\[ C = \frac{1}{25000 \times 10} = \frac{1}{250000} \]

### Step 7: Calculate the Final Value of \(C\)
Calculating \(C\):
\[ C = 4 \times 10^{-6} \, \text{F} = 4 \, \mu\text{F} \]

### Final Answer
The value of capacitance to maximize the average power is:
\[ C = 4 \, \mu\text{F} \]

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