A `100 V` a.c. source of frequency `500 Hz` is connected to a `LCR` circuit with `L = 8.1` millihenry, `C = 12.5 mu F` and `R = 10` ohm, all connected in series. What is the potential difference across the resistance?

Options

A. 25 V
B. 50 V
C. 75 V
D. 100 V

Answer

To find the potential difference across the resistance in the given LCR circuit, we can follow these steps:

### Step 1: Calculate the inductive reactance (XL)
The inductive reactance (XL) is given by the formula:
\[ X_L = \omega L \]
where \(\omega = 2\pi f\) and \(L\) is the inductance.

Given:
- \(f = 500 \, \text{Hz}\)
- \(L = 8.1 \, \text{mH} = 8.1 \times 10^{-3} \, \text{H}\)

First, calculate \(\omega\):
\[ \omega = 2\pi \times 500 = 1000\pi \approx 3141.59 \, \text{rad/s} \]

Now, calculate \(X_L\):
\[ X_L = 3141.59 \times 8.1 \times 10^{-3} \approx 25.4 \, \Omega \]

### Step 2: Calculate the capacitive reactance (XC)
The capacitive reactance (XC) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
where \(C\) is the capacitance.

Given:
- \(C = 12.5 \, \mu F = 12.5 \times 10^{-6} \, F\)

Now, calculate \(X_C\):
\[ X_C = \frac{1}{3141.59 \times 12.5 \times 10^{-6}} \approx 25.4 \, \Omega \]

### Step 3: Determine the impedance (Z)
In a series LCR circuit, the total impedance \(Z\) is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Since \(X_L \approx X_C\), we have:
\[ Z = R = 10 \, \Omega \]

### Step 4: Calculate the current (I)
Using Ohm's law, the current \(I\) can be calculated as:
\[ I = \frac{V_0}{Z} \]
where \(V_0\) is the voltage of the AC source.

Given:
- \(V_0 = 100 \, V\)

Substituting the values:
\[ I = \frac{100}{10} = 10 \, A \]

### Step 5: Calculate the potential difference across the resistance (VR)
The potential difference across the resistance \(V_R\) is given by:
\[ V_R = I \times R \]

Substituting the values:
\[ V_R = 10 \times 10 = 100 \, V \]

### Final Answer
The potential difference across the resistance is \(100 \, V\).

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