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 impedance of the circuit is unchanged
D. Total impedance of the circuit increases

Explanation

To solve the problem, we need to analyze the behavior of the circuit as the angular frequency \((\omega)\) is increased. The circuit consists of a capacitor \(C\) and a resistor \(R\) in series with an AC voltage source of fixed amplitude \(V_0\).

### Step-by-Step Solution:

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

**Hint**: Remember that impedance combines resistance and reactance (due to capacitance) in a right triangle.

2. **Effect of Increasing \(\omega\)**:
As \(\omega\) increases, the term \(\frac{1}{\omega C}\) decreases. This means that the capacitive reactance \(X_C = \frac{1}{\omega C}\) decreases. Therefore, the total impedance \(Z\) will also decrease because:
\[ Z = \sqrt{R^2 + X_C^2} \quad \text{(where \(X_C = \frac{1}{\omega C}\))} \]

**Hint**: Consider how changes in frequency affect capacitive reactance.

3. **Calculating RMS Current**:
The RMS current \(I_{\text{rms}}\) in the circuit can be calculated using Ohm's law for AC circuits:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} \]
Since \(V_{\text{rms}} = \frac{V_0}{\sqrt{2}}\) for a fixed amplitude \(V_0\), we can express it as:
\[ I_{\text{rms}} = \frac{V_0/\sqrt{2}}{Z} \]
As \(Z\) decreases (due to increasing \(\omega\)), \(I_{\text{rms}}\) will increase.

**Hint**: Remember that a decrease in impedance leads to an increase in current.

4. **Effect on the Bulb**:
The brightness of the bulb is directly related to the power dissipated in the resistor, which can be expressed as:
\[ P = I_{\text{rms}}^2 R \]
Since \(I_{\text{rms}}\) increases as \(\omega\) increases, the power \(P\) will also increase, causing the bulb to glow brighter.

**Hint**: Consider how power is related to current and resistance.

5. **Conclusion**:
Therefore, when the angular frequency \((\omega)\) is increased, the bulb will glow brighter.

### Final Answer:
The correct option is: **The bulb glows brighter.**