When a voltage `V = V_(0)` cos `omegat` is applied across a resistor of resistance R, the average power dissipated per cycle in the resistor is given by

Options

A. `(V_(0))/(sqrt(2)R)`
B. `(V_(0))/(sqrt(2)omega R)`
C. `(V_(0)^(2))/(2R)`
D. `(V_(0)^(2))/(2omegaR)`

Answer

To find the average power dissipated per cycle in a resistor when a voltage \( V = V_0 \cos(\omega t) \) is applied, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Voltage and Current:**
The voltage across the resistor is given by:
\[ V(t) = V_0 \cos(\omega t) \]
The current \( I(t) \) through the resistor \( R \) can be calculated using Ohm's law:
\[ I(t) = \frac{V(t)}{R} = \frac{V_0 \cos(\omega t)}{R} \]

2. **Calculate Instantaneous Power:**
The instantaneous power \( P(t) \) dissipated in the resistor is given by:
\[ P(t) = V(t) \cdot I(t) = V_0 \cos(\omega t) \cdot \frac{V_0 \cos(\omega t)}{R} = \frac{V_0^2 \cos^2(\omega t)}{R} \]

3. **Average Power Over One Cycle:**
To find the average power \( P_{avg} \) over one complete cycle, we need to integrate \( P(t) \) over one period \( T \) and then divide by \( T \):
\[ P_{avg} = \frac{1}{T} \int_0^T P(t) \, dt = \frac{1}{T} \int_0^T \frac{V_0^2 \cos^2(\omega t)}{R} \, dt \]

4. **Substituting the Period:**
The period \( T \) of the cosine function is given by:
\[ T = \frac{2\pi}{\omega} \]
Thus, we can substitute \( T \) into the average power equation:
\[ P_{avg} = \frac{1}{\frac{2\pi}{\omega}} \int_0^{\frac{2\pi}{\omega}} \frac{V_0^2 \cos^2(\omega t)}{R} \, dt \]

5. **Using the Identity for Cosine Squared:**
We can use the trigonometric identity:
\[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \]
Therefore:
\[ P(t) = \frac{V_0^2}{R} \cdot \frac{1 + \cos(2\omega t)}{2} \]

6. **Integrating:**
Now we can integrate:
\[ P_{avg} = \frac{1}{\frac{2\pi}{\omega}} \int_0^{\frac{2\pi}{\omega}} \frac{V_0^2}{2R} (1 + \cos(2\omega t)) \, dt \]
The integral of \( \cos(2\omega t) \) over one complete cycle is zero, so we only need to integrate the constant term:
\[ P_{avg} = \frac{1}{\frac{2\pi}{\omega}} \cdot \frac{V_0^2}{2R} \cdot \int_0^{\frac{2\pi}{\omega}} 1 \, dt \]
The integral of 1 over the period \( T \) is simply \( T \):
\[ P_{avg} = \frac{1}{\frac{2\pi}{\omega}} \cdot \frac{V_0^2}{2R} \cdot \frac{2\pi}{\omega} \]

7. **Final Simplification:**
This simplifies to:
\[ P_{avg} = \frac{V_0^2}{2R} \]

### Conclusion:
The average power dissipated per cycle in the resistor is:
\[ P_{avg} = \frac{V_0^2}{2R} \]