In an circuit, V and I are given by `V=150sin(150t) V` and `I =150sin(150t+(pi)/(3))A`. The power dissipated in the circuit is

Options

A. 106 W
B. 150 W
C. 5625 W
D. zero

Explanation

To find the power dissipated in the given circuit, we will follow these steps:

### Step 1: Identify the peak voltage and peak current
The voltage \( V \) and current \( I \) are given by:
- \( V = 150 \sin(150t) \)
- \( I = 150 \sin(150t + \frac{\pi}{3}) \)

From the voltage equation, we can identify the peak voltage \( V_0 \):
\[ V_0 = 150 \, \text{V} \]

From the current equation, we can identify the peak current \( I_0 \):
\[ I_0 = 150 \, \text{A} \]

### Step 2: Identify the phase difference
The phase difference \( \phi \) between the voltage and current can be extracted from the current equation:
\[ \phi = \frac{\pi}{3} \]

### Step 3: Use the formula for average power in an AC circuit
The average power \( P \) dissipated in an AC circuit is given by the formula:
\[ P = \frac{1}{2} V_0 I_0 \cos(\phi) \]

### Step 4: Substitute the values into the power formula
Now we substitute the values we found into the power formula:
\[ P = \frac{1}{2} \times 150 \times 150 \times \cos\left(\frac{\pi}{3}\right) \]

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

### Step 6: Substitute \( \cos\left(\frac{\pi}{3}\right) \) into the power equation
Now substituting \( \cos\left(\frac{\pi}{3}\right) \) into the equation:
\[ P = \frac{1}{2} \times 150 \times 150 \times \frac{1}{2} \]

### Step 7: Simplify the expression
Calculating the expression:
\[ P = \frac{1}{2} \times 150 \times 150 \times \frac{1}{2} = \frac{150 \times 150}{4} = \frac{22500}{4} = 5625 \, \text{W} \]

### Final Answer
The power dissipated in the circuit is:
\[ P = 5625 \, \text{W} \]

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