Question

The instantaneous values of alternating current and voltages in a circuit given as
$$ i = \frac{1}{\sqrt{2}} \sin(100\pi t) \text{ amp} $$ $$ e = \frac{1}{\sqrt{2}} \sin(100\pi t + \pi/3) \text{ volt} $$ The average power (in watts) consumed in the circuit is

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{\sqrt{3}}{2} \)
• D. \( \frac{1}{8} \)

Hint:

When calculating average power in AC circuits, use the formula \( P = \frac{1}{2} V_{\text{rms}} I_{\text{rms}} \), where \( V_{\text{rms}} \) and \( I_{\text{rms}} \) are the RMS values of voltage and current.

Answer 1

The Correct Option is D

The problem involves finding the average power consumed in an AC circuit given the instantaneous values of current and voltage. The expressions provided are:

Current: \( i = \frac{1}{\sqrt{2}} \sin(100\pi t) \text{ amp} \)

Voltage: \( e = \frac{1}{\sqrt{2}} \sin(100\pi t + \pi/3) \text{ volt} \)

The average power \( P \) consumed in an AC circuit is given by:

\( P = I_{\text{rms}} V_{\text{rms}} \cos\phi \)

Where:

• \( I_{\text{rms}} \) is the root mean square of the current.
• \( V_{\text{rms}} \) is the root mean square of the voltage.
• \( \phi \) is the phase difference between current and voltage.

From the expressions:

• \( I_{\text{rms}} = \frac{1}{\sqrt{2}} \)
• \( V_{\text{rms}} = \frac{1}{\sqrt{2}} \)

The phase difference \( \phi \) is \( \pi/3 \).

Substituting these values into the average power formula:

\( P = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) \cos\left(\frac{\pi}{3}\right) \)

We know that \( \cos(\pi/3) = \frac{1}{2} \). Thus:

\( P = \left(\frac{1}{\sqrt{2}}\right)^2 \times \frac{1}{2} \)

\( P = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \)

Therefore, the average power consumed in the circuit is \( \frac{1}{8} \) watts.

Answer 2

The expressions for the instantaneous current \( i \) and voltage \( e \) are given by:
\( i = \frac{1}{\sqrt{2}} \sin(100\pi t) \) amp
\( e = \frac{1}{\sqrt{2}} \sin(100\pi t + \pi/3) \) volt
To find the average power \( P \) consumed in the circuit, use the formula:
\( P = I_{\text{rms}} \cdot V_{\text{rms}} \cdot \cos(\phi) \)
where \( I_{\text{rms}} \) and \( V_{\text{rms}} \) are the root mean square values of the current and voltage respectively, and \( \phi \) is the phase difference.
Since the peak values are \( \frac{1}{\sqrt{2}} \), the RMS values are:
\( I_{\text{rms}} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \)
\( V_{\text{rms}} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \)
The phase difference \( \phi = \frac{\pi}{3} \)
Thus, the power factor \( \cos(\phi) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)
Now substitute back into the formula:
\( P = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} \text{ watts} \)