If the frequency of applied source is increased, how will brightness change?

Answer

To solve the problem of how brightness changes when the frequency of the applied source is increased, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Circuit Configuration**:
- The circuit consists of a bulb (which represents a load) connected in series with an inductor (L) and a resistor (R). The brightness of the bulb is related to the power consumed in the circuit.

2. **Identify the Impedance**:
- The total impedance (Z) in the circuit can be expressed as:
\[ Z = \sqrt{R^2 + X_L^2} \]
- Here, \(X_L\) is the inductive reactance, which is given by:
\[ X_L = \omega L = 2\pi f L \]
- Where \(f\) is the frequency of the applied source.

3. **Analyze the Effect of Increasing Frequency**:
- When the frequency \(f\) increases, the inductive reactance \(X_L\) also increases because \(X_L\) is directly proportional to frequency:
\[ X_L \propto f \]
- As a result, the total impedance \(Z\) of the circuit increases:
\[ Z = \sqrt{R^2 + (2\pi f L)^2} \]

4. **Relate Impedance to Power**:
- The power (P) consumed in the circuit can be expressed as:
\[ P = \frac{V^2}{Z} \]
- Here, \(V\) is the voltage across the circuit. Since \(Z\) is in the denominator, as \(Z\) increases, the power \(P\) decreases.

5. **Conclude the Effect on Brightness**:
- The brightness of the bulb is directly proportional to the power consumed. Therefore, if the power decreases due to the increase in impedance, the brightness of the bulb will also decrease.

### Final Answer:
If the frequency of the applied source is increased, the brightness of the bulb will decrease.