A bulb and a conender are connected in series with an A.C source. On increasing the frequency of the source its brightness will

Options

A. increase
B. decrease
C. sometimes increase and sometimes decrease
D. neigther increase nor decrease .

Explanation

To solve the problem step by step, we need to analyze how the brightness of the bulb changes when the frequency of the AC source is increased.

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a bulb (which can be modeled as a resistor, R) and a capacitor (condenser) connected in series with an AC source.

2. **Power and Brightness Relation**:
- The brightness of the bulb is related to the power consumed by it. The power (P) in an AC circuit can be expressed as:
\[ P = \frac{V^2}{R} \]
- Here, V is the voltage across the bulb, and R is the resistance of the bulb.

3. **Impedance in the Circuit**:
- The total impedance (Z) in the circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
- Where \(X_L\) is the inductive reactance (\(X_L = \omega L\)) and \(X_C\) is the capacitive reactance (\(X_C = \frac{1}{\omega C}\)).

4. **Effect of Increasing Frequency**:
- When the frequency of the AC source is increased, the angular frequency (\(\omega\)) increases.
- This leads to an increase in \(X_L\) (since \(X_L = \omega L\)) and a decrease in \(X_C\) (since \(X_C = \frac{1}{\omega C}\)).
- Therefore, the difference \(X_L - X_C\) increases as frequency increases.

5. **Change in Impedance**:
- Since \(X_L\) increases and \(X_C\) decreases, the overall impedance \(Z\) increases:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
- As \(Z\) increases, the current (I) in the circuit decreases (Ohm's Law: \(I = \frac{V}{Z}\)).

6. **Voltage Across the Bulb**:
- The voltage across the bulb can be expressed as:
\[ V = I \cdot R \]
- Since \(I\) decreases due to the increase in \(Z\), the voltage \(V\) across the bulb will also decrease.

7. **Brightness of the Bulb**:
- Since the brightness of the bulb is proportional to \(V^2\), if \(V\) decreases, the brightness of the bulb will also decrease.

### Conclusion:
- Therefore, on increasing the frequency of the AC source, the brightness of the bulb will **decrease**.