In a series LCR circuit, an alternating emf (V) and current (I) are given by the equation `V =V_0sinomegat,I=I_0sin(omegat+pi/3)` The average power dissipated in the circuit over a cycle of AC is

Options

A. `(V_0I_0)/2`
B. `(V_0I_0)/4`
C. `sqrt3/2V_0I_0`
D. Zero

Answer

To find the average power dissipated in a series LCR circuit with the given alternating emf (V) and current (I), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given equations**:
- The voltage across the circuit is given by:
\[ V = V_0 \sin(\omega t) \]
- The current through the circuit is given by:
\[ I = I_0 \sin(\omega t + \frac{\pi}{3}) \]

2. **Determine the phase difference (φ)**:
- The phase difference between the voltage and the current can be identified from the equations. The voltage is in phase with \(\sin(\omega t)\), while the current leads by \(\frac{\pi}{3}\). Thus, the phase difference \(φ\) is:
\[ φ = \frac{\pi}{3} \]

3. **Use the formula for average power (P)**:
- The average power \(P\) in an AC circuit is given by:
\[ P = \frac{V_0 I_0}{2} \cos(φ) \]

4. **Substitute the values into the power formula**:
- Substitute \(φ = \frac{\pi}{3}\) into the power formula:
\[ P = \frac{V_0 I_0}{2} \cos\left(\frac{\pi}{3}\right) \]

5. **Calculate \(\cos\left(\frac{\pi}{3}\right)\)**:
- We know that:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

6. **Substitute \(\cos\left(\frac{\pi}{3}\right)\) back into the power formula**:
- Thus, we have:
\[ P = \frac{V_0 I_0}{2} \cdot \frac{1}{2} \]
- Simplifying this gives:
\[ P = \frac{V_0 I_0}{4} \]

### Final Answer:
The average power dissipated in the circuit over a cycle of AC is:
\[ P = \frac{V_0 I_0}{4} \]