An alternating potential V = `underset(o)(V) sin omega`t is applied across a circuit. As a result the current,
I = `underset(o)(I)sin(omegat - pi/2)` flows in it. The power consecutive in the circuit per cycle is

Options

A. zero
B. 0.5`V_(0)` and `I_(0)`
C. 0.707`V_(0)` and `I_(0)`
D. 1.414`V_(0)` and `I_(0)`

Answer

To solve the problem, we need to find the power consumed in the circuit when an alternating potential is applied. Here are the steps to derive the solution:

### Step 1: Identify the given quantities
We have:
- The voltage across the circuit: \( V = V_0 \sin(\omega t) \)
- The current flowing through the circuit: \( I = I_0 \sin(\omega t - \frac{\pi}{2}) \)

### Step 2: Determine the phase difference
The phase of the voltage is \( \omega t \) and the phase of the current is \( \omega t - \frac{\pi}{2} \). The phase difference \( \phi \) between the voltage and the current can be calculated as:
\[ \phi = \text{Phase of Voltage} - \text{Phase of Current} = \omega t - \left(\omega t - \frac{\pi}{2}\right) = \frac{\pi}{2} \]

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

### Step 4: Use the power formula
The average power \( P \) consumed in an AC circuit is given by the formula:
\[ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi) \]

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

### Step 6: Evaluate the cosine term
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ P = \left(\frac{V_0}{\sqrt{2}}\right) \cdot \left(\frac{I_0}{\sqrt{2}}\right) \cdot 0 = 0 \]

### Conclusion
The average power consumed in the circuit per cycle is:
\[ P = 0 \]

### Final Answer
The power consumed in the circuit per cycle is **0**.

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