A voltage of peak value 283 V and varying frequency is applied to series LCR combination in which `R=3Omega, L=25 mH` and `C=400 mu F`. Then the frequency (in Hz) of the source at which maximum power is dissipated in the above is

Options

A. `51.5`
B. `50.7`
C. `51.1`
D. `50.3`

Explanation

To find the frequency at which maximum power is dissipated in a series LCR circuit, we need to calculate the resonant frequency. The resonant frequency \( f_0 \) for an LCR circuit is given by the formula:

\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]

where:
- \( L \) is the inductance in henries (H)
- \( C \) is the capacitance in farads (F)

### Step 1: Convert the given values to standard units
- Inductance \( L = 25 \, \text{mH} = 25 \times 10^{-3} \, \text{H} \)
- Capacitance \( C = 400 \, \mu\text{F} = 400 \times 10^{-6} \, \text{F} \)

### Step 2: Substitute the values into the resonant frequency formula
Now, substitute the values of \( L \) and \( C \) into the formula:

\[ f_0 = \frac{1}{2\pi\sqrt{(25 \times 10^{-3})(400 \times 10^{-6})}} \]

### Step 3: Calculate the product \( LC \)
First, calculate \( LC \):

\[ LC = (25 \times 10^{-3}) \times (400 \times 10^{-6}) = 10 \times 10^{-6} = 10^{-5} \, \text{H}\cdot\text{F} \]

### Step 4: Calculate the square root of \( LC \)
Now calculate \( \sqrt{LC} \):

\[ \sqrt{LC} = \sqrt{10^{-5}} = 10^{-2.5} = 0.003162 \, \text{(approximately)} \]

### Step 5: Calculate \( f_0 \)
Now substitute \( \sqrt{LC} \) back into the resonant frequency formula:

\[ f_0 = \frac{1}{2\pi \times 0.003162} \]

Calculating this gives:

\[ f_0 \approx \frac{1}{0.019869} \approx 50.3 \, \text{Hz} \]

### Conclusion
The frequency at which maximum power is dissipated in the LCR circuit is approximately:

\[ \boxed{50.3 \, \text{Hz}} \]