Question:medium

In a given circuit, the instantaneous values of alternating voltage and current are $V = 5 \sin(80\pi t + \pi)$ volts and $I = 5 \sin(80\pi t)$ amperes respectively. Find the average power consumed in the circuit.

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When two electrical wave parameters are completely out of phase by $\pi$ radians ($180^\circ$), the power factor $\cos\phi$ drops to $-1$. This means the system power expression simply becomes $-\frac{V_0 I_0}{2} = -\frac{0.25}{2} = -0.125\text{ W}$.
Updated On: May 20, 2026
  • $0.0625\text{ W}$
  • $1.25\text{ W}$
  • $0.625\text{ W}$
  • $-0.125\text{ W}$
Show Solution

The Correct Option is D

Solution and Explanation

Understanding the Concept: The average electrical power ($P_{\text{avg}}$) consumed in an alternating current circuit depends on the root-mean-square (rms) voltage, rms current, and the phase difference ($\phi$) between them: \[ P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi = \frac{V_0}{\sqrt{2}} \frac{I_0}{\sqrt{2}} \cos\phi = \frac{V_0 I_0}{2} \cos\phi \]
Step 1: Identify peak values and find the relative phase angle difference ($\phi$).
From the given equations:
Peak Voltage, $V_0 = 0.5\text{ V}$
Peak Current, $I_0 = 0.5\text{ A}$
Phase configuration matching, $\phi_V = 80\pi t + \pi$ and $\phi_I = 80\pi t$
The net phase difference $\phi$ is: \[ \phi = \phi_V - \phi_I = (80\pi t + \pi) - 80\pi t = \pi \quad (180^\circ) \]
Step 2: Substitute values into the average power formula.
\[ P_{\text{avg}} = \frac{0.5 \times 0.5}{2} \times \cos(\pi) \] We know that $\cos(\pi) = -1$: \[ P_{\text{avg}} = \frac{0.25}{2} \times (-1) = -0.125\text{ W} \] (Note: A negative power value indicates that the phase-inverted source configuration is structurally returning power back to the primary line environment network).
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