An AC voltage source of variable angular frequency `(omega)` and fixed amplitude `V_(0)` is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When `(omega)` is increased

Options

A. the bulb glows dimmer
B. the bulb glows brighter
C. total impedence of the circuit is unchanged
D. total impedence of the circuit increases

Explanation

To solve the problem, we need to analyze the behavior of the circuit as the angular frequency \( \omega \) of the AC voltage source is increased. The circuit consists of a capacitor \( C \) and a resistor \( R \) connected in series.

### Step-by-Step Solution:

1. **Understand the Impedance in the Circuit**:
The impedance \( Z \) of a series circuit containing a resistor \( R \) and a capacitor \( C \) is given by the formula:
\[ Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \]
Here, \( \omega \) is the angular frequency of the AC source.

2. **Analyze the Effect of Increasing \( \omega \)**:
As \( \omega \) increases, the term \( \frac{1}{\omega C} \) decreases because it is inversely proportional to \( \omega \). This means that the capacitive reactance \( X_C = \frac{1}{\omega C} \) decreases.

3. **Calculate the New Impedance**:
Since \( X_C \) decreases, the overall impedance \( Z \) of the circuit will also decrease:
\[ Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \quad \text{(with decreasing } \frac{1}{\omega C}\text{)} \]
As \( \frac{1}{\omega C} \) becomes smaller, \( Z \) approaches \( R \).

4. **Determine the Current in the Circuit**:
The current \( I \) flowing through the circuit can be expressed using Ohm's law as:
\[ I = \frac{V_0}{Z} \]
Since \( Z \) decreases as \( \omega \) increases, the current \( I \) will increase.

5. **Effect on the Bulb**:
The brightness of the bulb is proportional to the power dissipated in the resistor \( R \), which can be calculated as:
\[ P = I^2 R \]
Since \( I \) increases with decreasing \( Z \), the power \( P \) also increases.

6. **Conclusion**:
Therefore, as \( \omega \) is increased, the bulb will glow brighter due to the increase in current and power dissipated in the resistor.

### Final Answer:
As the angular frequency \( \omega \) is increased, the bulb will glow brighter.