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 1: Identify the Instantaneous Current
Given the voltage across the resistor:
\[ V(t) = V_0 \cos(\omega t) \]
The current \( I(t) \) through the resistor \( R \) can be expressed using Ohm's law:
\[ I(t) = \frac{V(t)}{R} = \frac{V_0 \cos(\omega t)}{R} \]

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

### Step 3: Find the Average Power Over One Cycle
To find the average power \( P_{\text{avg}} \) over one complete cycle, we need to integrate the instantaneous power over one period \( T \) and then divide by \( T \):
\[ P_{\text{avg}} = \frac{1}{T} \int_0^T P(t) \, dt = \frac{1}{T} \int_0^T \frac{V_0^2 \cos^2(\omega t)}{R} \, dt \]
Since \( R \) and \( V_0^2 \) are constants, we can factor them out:
\[ P_{\text{avg}} = \frac{V_0^2}{R} \cdot \frac{1}{T} \int_0^T \cos^2(\omega t) \, dt \]

### Step 4: Evaluate the Integral
The integral of \( \cos^2(\omega t) \) over one period can be evaluated using the identity:
\[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \]
Thus,
\[ \int_0^T \cos^2(\omega t) \, dt = \int_0^T \frac{1 + \cos(2\omega t)}{2} \, dt = \frac{1}{2} \left( \int_0^T 1 \, dt + \int_0^T \cos(2\omega t) \, dt \right) \]
The first integral evaluates to \( T \), and the second integral evaluates to zero over one complete cycle:
\[ \int_0^T \cos(2\omega t) \, dt = 0 \]
Thus,
\[ \int_0^T \cos^2(\omega t) \, dt = \frac{1}{2} T \]

### Step 5: Substitute Back to Find Average Power
Now substituting back into the average power equation:
\[ P_{\text{avg}} = \frac{V_0^2}{R} \cdot \frac{1}{T} \cdot \frac{1}{2} T = \frac{V_0^2}{2R} \]

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