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 will follow these steps:

### Step 1: Identify the given values
- Voltage: \( V(t) = 50 \sin(50t) \) volts
- Current: \( I(t) = 100 \sin(50t + \frac{\pi}{3}) \) mA

### Step 2: Determine the maximum values of voltage and current
- The maximum voltage \( V_0 = 50 \) volts
- The maximum current \( I_0 = 100 \) mA = \( 100 \times 10^{-3} \) A = \( 0.1 \) A

### Step 3: Calculate the RMS values of voltage and current
- The RMS value of voltage \( V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{50}{\sqrt{2}} \) volts
- The RMS value of current \( I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} \) A

### Step 4: Determine the phase difference
- The phase difference \( \phi = \frac{\pi}{3} \)

### Step 5: Calculate the power factor
- The power factor \( \cos(\phi) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)

### Step 6: Calculate the average power dissipated in the circuit
Using the formula for power in an AC circuit:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]
Substituting the values:
\[ P = \left(\frac{50}{\sqrt{2}}\right) \cdot \left(\frac{0.1}{\sqrt{2}}\right) \cdot \frac{1}{2} \]

### Step 7: Simplify the expression
\[ P = \frac{50 \cdot 0.1}{2} \cdot \frac{1}{2} = \frac{5}{2} = 2.5 \text{ watts} \]

### Final Answer
The power dissipated in the circuit is \( 2.5 \) watts.

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