In a purely resistive circuit, the average power is _________ the peak power.

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Examoogle · Accepted Answer

one-half of

Explanation: In a purely resistive AC circuit, the average power dissipated is exactly one-half of the peak (maximum) instantaneous power. This occurs because the power waveform is a shifted sine wave oscillating at twice the frequency of the voltage or current, oscillating between zero and the peak value.

Working Principle: The instantaneous power $p(t)$ is the product of instantaneous voltage $v(t) = V_m \sin(\omega t)$ and current $i(t) = I_m \sin(\omega t)$. Thus, $p(t) = V_m I_m \sin^2(\omega t) = V_m I_m \left(\frac{1 - \cos(2\omega t)}{2}ight)$. The term $V_m I_m$ represents the peak power ($P_{peak}$), and the average value of this oscillating term over a cycle is $\frac{V_m I_m}{2}$.

Key Formulas: $P_{avg} = \frac{V_m I_m}{2} = I_{rms}^2 R$; $P_{peak} = V_m I_m = 2 P_{avg}$

Key Points: In purely resistive circuits, voltage and current are in phase.; The instantaneous power varies sinusoidally at twice the supply frequency.; The peak power $V_m I_m$ is achieved at the peak of the sine wave cycle.; The average power is the mean value of the squared signal over time.

Advantages: Simplicity in calculating power dissipation.; No phase shift implies unity power factor.

Disadvantages: Peak power stress on insulation components.; High instantaneous values can cause localized heating.

Applications: Heating elements (coils, toasters).; Incandescent lighting.; Standard resistor load testing.

Additional Info: For sinusoidal sources, $P_{peak} = 2 	imes P_{avg}$.. Option A is incorrect as it describes the inverse relationship. Option C and D are mathematically invalid for sinusoidal signals.

Answer: B is correct — The average power in a purely resistive circuit is always half of the peak instantaneous power.

Exam Tip: Remember that for purely reactive elements (inductors/capacitors), average power is zero because the power cycles symmetrically above and below the zero axis.

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