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. 150W
C. 5625 W
D. zero

Explanation

To find the power dissipated in the circuit given the voltage and current equations, we can follow these steps:

### Step 1: Identify the RMS values of voltage and current
The voltage \( V(t) \) is given by:
\[ V(t) = 150 \sin(150t) \, V \]
The peak voltage \( V_0 \) is 150 V. The RMS (Root Mean Square) value of voltage is calculated as:
\[ V_{RMS} = \frac{V_0}{\sqrt{2}} = \frac{150}{\sqrt{2}} = 75\sqrt{2} \, V \]

The current \( I(t) \) is given by:
\[ I(t) = 150 \sin\left(150t + \frac{\pi}{3}\right) \, A \]
The peak current \( I_0 \) is also 150 A. The RMS value of current is:
\[ I_{RMS} = \frac{I_0}{\sqrt{2}} = \frac{150}{\sqrt{2}} = 75\sqrt{2} \, A \]

### Step 2: Determine the phase difference
From the equations, we can see that the current leads the voltage by an angle of \( \phi = \frac{\pi}{3} \).

### Step 3: Calculate the power dissipated
The average power \( P \) dissipated in the circuit can be calculated using the formula:
\[ P = V_{RMS} \times I_{RMS} \times \cos(\phi) \]
Substituting the values we found:
\[ P = (75\sqrt{2}) \times (75\sqrt{2}) \times \cos\left(\frac{\pi}{3}\right) \]
We know that \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ P = (75\sqrt{2}) \times (75\sqrt{2}) \times \frac{1}{2} \]
Calculating \( (75\sqrt{2})^2 \):
\[ (75\sqrt{2})^2 = 75^2 \times 2 = 5625 \times 2 = 11250 \]
Thus,
\[ P = \frac{11250}{2} = 5625 \, W \]

### Final Answer:
The power dissipated in the circuit is \( 5625 \, W \).

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