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 solve the problem of finding the average power dissipated in a series LCR circuit given the equations for alternating emf (V) and current (I), we can follow these steps:

### Step 1: Identify the given equations
The equations provided are:
- \( V = V_0 \sin(\omega t) \)
- \( I = I_0 \sin(\omega t + \frac{\pi}{3}) \)

### Step 2: Determine the phase difference (\( \phi \))
The phase difference between the voltage and current can be calculated as follows:
\[ \phi = \left( \omega t + \frac{\pi}{3} \right) - \omega t = \frac{\pi}{3} \]

### Step 3: Calculate the RMS values of voltage and current
The root mean square (RMS) values for voltage and current are given by:
\[ V_{rms} = \frac{V_0}{\sqrt{2}}, \quad I_{rms} = \frac{I_0}{\sqrt{2}} \]

### Step 4: Use the formula for average power
The average power (\( P_{avg} \)) dissipated in the circuit is given by:
\[ P_{avg} = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]

### Step 5: Substitute the values into the power formula
Substituting the RMS values and the phase difference into the power formula:
\[ P_{avg} = \left( \frac{V_0}{\sqrt{2}} \right) \cdot \left( \frac{I_0}{\sqrt{2}} \right) \cdot \cos\left(\frac{\pi}{3}\right) \]

### Step 6: Calculate \( \cos\left(\frac{\pi}{3}\right) \)
The value of \( \cos\left(\frac{\pi}{3}\right) \) is:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 7: Substitute and simplify
Now substituting \( \cos\left(\frac{\pi}{3}\right) \) into the equation:
\[ P_{avg} = \left( \frac{V_0}{\sqrt{2}} \cdot \frac{I_0}{\sqrt{2}} \cdot \frac{1}{2} \right) \]
\[ P_{avg} = \frac{V_0 I_0}{2 \cdot 2} = \frac{V_0 I_0}{4} \]

### Final Answer
The average power dissipated in the LCR circuit is:
\[ P_{avg} = \frac{V_0 I_0}{4} \]

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