In a series lags the applied emf. The rate at which energy is dissipated in the resistor, can be increased by

Options

A. Decreasing the capacitance and making no other change
B. Increasing the capacitance and making no other change
C. Increasing the inductance and making no other change
D. Increasing the driving frequency and making no other change

Answer

To solve the problem of how to increase the rate at which energy is dissipated in a resistor in a series LCR circuit where the current lags the applied emf, we can follow these steps:

### Step 1: Understand the Power Dissipation in a Resistor
The power dissipated in a resistor (P) in an AC circuit can be expressed as:
\[ P = I^2 R \]
where \( I \) is the current through the resistor and \( R \) is the resistance.

### Step 2: Identify Factors Affecting Current
The current in an LCR circuit is affected by the impedance (Z) of the circuit, which is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where:
- \( X_L = \omega L \) (inductive reactance)
- \( X_C = \frac{1}{\omega C} \) (capacitive reactance)
- \( \omega \) is the angular frequency of the AC source.

### Step 3: Analyze the Given Options
We need to analyze how each option affects the impedance and thus the current:

1. **Decreasing the Capacitance (C)**:
- If capacitance \( C \) decreases, \( X_C \) increases (since \( X_C = \frac{1}{\omega C} \)).
- This results in \( X_L - X_C \) decreasing, which leads to a decrease in impedance \( Z \).
- Since \( I = \frac{V}{Z} \), a decrease in \( Z \) results in an increase in current \( I \).
- Thus, power \( P = I^2 R \) increases.

2. **Increasing the Capacitance (C)**:
- If capacitance \( C \) increases, \( X_C \) decreases.
- This results in \( X_L - X_C \) increasing, which leads to an increase in impedance \( Z \).
- This results in a decrease in current \( I \), thus decreasing power \( P \).

3. **Increasing the Inductance (L)**:
- If inductance \( L \) increases, \( X_L \) increases.
- This results in \( X_L - X_C \) increasing, which leads to an increase in impedance \( Z \).
- This results in a decrease in current \( I \), thus decreasing power \( P \).

4. **Increasing the Driving Frequency (\( \omega \))**:
- If frequency \( \omega \) increases, \( X_L \) increases and \( X_C \) decreases.
- The net effect on \( X_L - X_C \) depends on the relative rates of change, but typically leads to an increase in impedance \( Z \).
- This results in a decrease in current \( I \), thus decreasing power \( P \).

### Conclusion
From the analysis, the only option that increases the power dissipation in the resistor is:

**Decreasing the capacitance and making no other changes.**

### Final Answer
The rate at which energy is dissipated in the resistor can be increased by **decreasing the capacitance and making no other changes.**

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