A student in a lab took a coil and connected it to a `12 VDC` source. He measures the steady state current in the circuit to be `4 A`. He then replaced the `12 VDC` source by a `12V,(omega=50rad/s)AC` source and observes that the reading in the `AC` ammeter is `2.4 A`. He then decides to connect a `2500 muF` capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).
Which of the following graph roughly matches the variation of current in the circuit (with the coil and capacitor connected in the series) when the angulr frequency is decreased from 50 rad/s to 25 rad/s?

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the behavior of the circuit when a capacitor is added in series with an inductor (coil) and how the current varies with a change in angular frequency. Here’s a step-by-step breakdown:

### Step 1: Understand the Circuit Components
- The circuit consists of a coil (inductor) and a capacitor connected in series.
- The initial measurements were taken with a DC source and then with an AC source at an angular frequency of \( \omega = 50 \, \text{rad/s} \).

### Step 2: Calculate the Impedance of the Circuit
1. **Impedance Formula**: For a series RLC circuit, the impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where \( R \) is the resistance, \( X_L \) is the inductive reactance, and \( X_C \) is the capacitive reactance.

2. **Given Values**:
- From the DC circuit, the resistance \( R \) can be calculated using Ohm's law:
\[ R = \frac{V}{I} = \frac{12 \, \text{V}}{4 \, \text{A}} = 3 \, \Omega \]
- The AC current is given as \( 2.4 \, \text{A} \) at \( \omega = 50 \, \text{rad/s} \).

3. **Calculate \( X_L \)**:
- The inductive reactance \( X_L \) can be calculated using:
\[ I = \frac{V}{Z} \implies Z = \frac{V}{I} = \frac{12 \, \text{V}}{2.4 \, \text{A}} = 5 \, \Omega \]
- Rearranging the impedance formula gives:
\[ 5 = \sqrt{3^2 + (X_L - X_C)^2} \]
- Solving for \( (X_L - X_C) \):
\[ 5^2 = 9 + (X_L - X_C)^2 \implies 25 - 9 = (X_L - X_C)^2 \implies (X_L - X_C)^2 = 16 \implies |X_L - X_C| = 4 \]

### Step 3: Calculate Capacitive Reactance
1. **Capacitive Reactance**:
- The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{\omega C} \]
- Given \( C = 2500 \, \mu F = 2500 \times 10^{-6} \, F \):
- At \( \omega = 50 \, \text{rad/s} \):
\[ X_C = \frac{1}{50 \times 2500 \times 10^{-6}} = \frac{1}{0.125} = 8 \, \Omega \]

### Step 4: Analyze the Effect of Decreasing Angular Frequency
1. **Decrease \( \omega \)**:
- When \( \omega \) decreases from \( 50 \, \text{rad/s} \) to \( 25 \, \text{rad/s} \), \( X_C \) increases:
\[ X_C = \frac{1}{25 \times 2500 \times 10^{-6}} = \frac{1}{0.05} = 20 \, \Omega \]

2. **Calculate New Impedance**:
- The new impedance \( Z \) when \( \omega = 25 \, \text{rad/s} \):
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
- Since \( X_L - X_C \) will now be \( 4 - 20 = -16 \):
\[ Z = \sqrt{3^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265} \approx 16.28 \, \Omega \]

### Step 5: Calculate the New Current
- The new current \( I \) can be calculated using:
\[ I = \frac{V}{Z} = \frac{12 \, \text{V}}{16.28 \, \Omega} \approx 0.737 \, A \]

### Step 6: Graphical Representation
- As the angular frequency decreases, the current decreases due to the increase in capacitive reactance, resulting in a higher impedance.
- The graph that represents the current variation as \( \omega \) decreases from \( 50 \, \text{rad/s} \) to \( 25 \, \text{rad/s} \) would show a decreasing trend in current.

### Conclusion
The correct graph that matches the variation of current in the circuit when the angular frequency is decreased from \( 50 \, \text{rad/s} \) to \( 25 \, \text{rad/s} \) is the one showing a decrease in current.