Question

In a circuit, voltage and current are given by $V = 10 \sin(\omega t + 30^\circ)$ and $i = 10 \sin(\omega t - 30^\circ)$. The power consumed in the circuit is ____.

• A. 100 watts
• B. 50 watts
• C. 25 watts
• D. 12.5 watts

Hint:

Be careful with the phase! Always subtract the smaller angle from the larger one or follow the order $(V_{\text{angle}} - I_{\text{angle}})$. Here, $+30^\circ$ and $-30^\circ$ are $60^\circ$ apart on a phasor diagram.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In AC circuits, the average power consumed depends on the RMS values of voltage and current and the phase difference between them (the power factor).

Step 2: Key Formula or Approach:
The average power $P$ is given by: \[ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi \] Where $V_{\text{rms}} = \frac{V_m}{\sqrt{2}}$, $I_{\text{rms}} = \frac{I_m}{\sqrt{2}}$, and $\phi$ is the phase difference.

Step 3: Detailed Explanation:
1. Identify Peak Values: From the equations, peak voltage $V_m = 10$ and peak current $I_m = 10$. 2. Calculate Phase Difference ($\phi$): \[ \phi = (\text{Phase of } V) - (\text{Phase of } i) \] \[ \phi = 30^\circ - (-30^\circ) = 60^\circ \] 3. Calculate Power: \[ P = \left( \frac{10}{\sqrt{2}} \right) \cdot \left( \frac{10}{\sqrt{2}} \right) \cdot \cos(60^\circ) \] \[ P = \frac{100}{2} \cdot \frac{1}{2} \] \[ P = 50 \cdot 0.5 = 25\text{ Watts} \]

Step 4: Final Answer:
The power consumed in the circuit is 25 watts.

Answer 2

Instead of first converting to RMS values and then multiplying, we can use the direct formula for average power in terms of peak values, \( P = \frac{1}{2} V_m I_m \cos\phi \), which comes from averaging the instantaneous power over one full cycle.

1. 100 watts: This would result from simply multiplying the peak voltage and peak current together, \( V_m I_m = 10 \times 10 = 100 \), while forgetting both the factor of \( \frac{1}{2} \) that comes from time-averaging a sinusoid and the phase difference between voltage and current entirely.
2. 50 watts: This would result from applying the \( \frac{1}{2} \) averaging factor, \( \frac{1}{2} \times 100 = 50 \), but forgetting to multiply by \( \cos\phi \), effectively assuming voltage and current are perfectly in phase, \( \phi = 0 \), which is not the case here.
3. 25 watts: The phase difference between \( V = 10\sin(\omega t + 30^\circ) \) and \( i = 10\sin(\omega t - 30^\circ) \) is \( \phi = 30^\circ - (-30^\circ) = 60^\circ \). Applying the full formula: \[ P = \frac{1}{2} V_m I_m \cos\phi = \frac{1}{2}(10)(10)\cos 60^\circ = 50 \times 0.5 = 25 \text{ W} \] This matches the correctly computed average power.
4. 12.5 watts: This would result from an additional, unwarranted halving of the correct result, for instance mistakenly applying the \( \frac{1}{2} \) averaging factor a second time, or misreading the phase angle combined incorrectly.

Using the peak-value power formula directly gives the same 25 W obtained through the RMS approach.

Therefore, the correct answer is 25 watts.