In an a.c. circuit V and I are given by V=50 sin50t volt and I = 100 sin`(50t + pi//3)` mA. The power dissipated in the circuit

Options

A. 2.5 W
B. 1.25 W
C. 5.0 W
D. 500 watt

Answer

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

### Step 1: Identify the given values
We have the voltage \( V \) and current \( I \) given as:
- \( V = 50 \sin(50t) \) volts
- \( I = 100 \sin(50t + \frac{\pi}{3}) \) mA

### Step 2: Determine the peak values of voltage and current
From the equations:
- The peak voltage \( V_0 = 50 \) volts
- The peak current \( I_0 = 100 \) mA = \( 0.1 \) A (since 1 mA = 0.001 A)

### Step 3: Calculate the phase difference
The phase difference \( \phi \) between the voltage and current can be determined from the current equation:
- \( I = 100 \sin(50t + \frac{\pi}{3}) \)
Here, the phase shift is \( \frac{\pi}{3} \) radians.

### Step 4: Use the formula for average power
The average power \( P \) dissipated in an AC circuit is given by the formula:
\[ P = \frac{V_0 I_0}{2} \cos(\phi) \]
Substituting the values we have:
- \( V_0 = 50 \) volts
- \( I_0 = 0.1 \) A
- \( \phi = \frac{\pi}{3} \)

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

### Step 6: Substitute values into the power formula
Now substituting the values into the power formula:
\[ P = \frac{50 \times 0.1}{2} \cos\left(\frac{\pi}{3}\right) \]
\[ P = \frac{50 \times 0.1}{2} \times \frac{1}{2} \]
\[ P = \frac{5}{2} \times \frac{1}{2} \]
\[ P = \frac{5}{4} \text{ watts} \]

### Step 7: Final result
Thus, the average power dissipated in the circuit is:
\[ P = 1.25 \text{ watts} \]

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