(a)If the current in a sinusoidally driven series RLC circuit leads the emf , would we increase or decrease the capacitance to increase the rate at which energy is supplied to the resistance ? (b) Would this change bring the resonant angular frequency of the circuit closer to the angular frequency of the emf or put it father away ?

Answer

To solve the given question step by step, we will analyze both parts (a) and (b) of the question regarding the RLC circuit.

### Step-by-Step Solution:

**(a)** We need to determine whether to increase or decrease the capacitance (C) in a sinusoidally driven series RLC circuit where the current leads the emf.

1. **Understanding the Circuit**: In a series RLC circuit, the total impedance (Z) is affected by the resistance (R), inductive reactance (XL), and capacitive reactance (XC). The relationship is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]

2. **Current Leading the EMF**: If the current leads the emf, it indicates that the circuit is capacitive. This means that the capacitive reactance (XC) is greater than the inductive reactance (XL):
\[ X_C > X_L \]

3. **Capacitive Reactance Formula**: The capacitive reactance is given by:
\[ X_C = \frac{1}{2\pi f C} \]
where \( f \) is the frequency of the AC source.

4. **Effect of Capacitance on Reactance**: Since \( X_C \) is inversely proportional to \( C \), decreasing \( C \) will increase \( X_C \). Therefore, to increase the rate at which energy is supplied to the resistance, we should **decrease the capacitance**.

5. **Conclusion for Part (a)**: To increase the rate at which energy is supplied to the resistance, we should **decrease the capacitance (C)**.

---

**(b)** Now we need to determine how this change in capacitance affects the resonant angular frequency of the circuit.

1. **Resonant Frequency Formula**: The resonant frequency (\( \omega_0 \)) of a series RLC circuit is given by:
\[ \omega_0 = \frac{1}{\sqrt{LC}} \]

2. **Effect of Decreasing Capacitance**: If we decrease the capacitance (C), the resonant frequency will increase because \( \omega_0 \) is inversely proportional to the square root of \( C \):
\[ \omega_0 \propto \frac{1}{\sqrt{C}} \]

3. **Comparison with Angular Frequency of EMF**: If the resonant frequency increases, it will move closer to the angular frequency of the emf (assuming the emf frequency is constant).

4. **Conclusion for Part (b)**: Decreasing the capacitance will bring the resonant angular frequency **closer** to the angular frequency of the emf.

---

### Final Answers:
(a) Decrease the capacitance (C) to increase the rate at which energy is supplied to the resistance.
(b) This change will bring the resonant angular frequency closer to the angular frequency of the emf.

---