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

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Examoogle · Accepted Answer

one-half of

Explanation: In a purely resistive circuit, the instantaneous power varies sinusoidally at twice the supply frequency. The average power, calculated over a full cycle, is exactly half of the peak (maximum) instantaneous power.

Working Principle: The instantaneous power $p(t) = v(t)i(t) = V_m\sin(\omega t) \cdot I_m\sin(\omega t) = V_m I_m \sin^2(\omega t)$. Using the identity $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$, we get $p(t) = \frac{V_m I_m}{2} - \frac{V_m I_m}{2}\cos(2\omega t)$. The first term represents the average power $P_{avg} = \frac{V_m I_m}{2}$, which is clearly half of the peak power $P_{max} = V_m I_m$.

Key Formulas: $P_{avg} = \frac{V_m I_m}{2}$ — Average power in resistive AC circuits; $p(t) = V_m I_m \sin^2(\omega t)$ — Instantaneous power

Key Points: Power in purely resistive circuits is always non-negative.; The power frequency is $2f$, where $f$ is the supply frequency.; Peak power occurs when voltage and current are simultaneously at their maximum values.

Advantages: Simple mathematical relationship; No phase shift interference

Disadvantages: Only applicable to purely resistive loads; Does not account for reactive components

Applications: Heaters and incandescent lighting; Basic AC circuit power analysis

Additional Info: For AC circuits, $P_{avg} = V_{rms} I_{rms} = \frac{V_m}{\sqrt{2}} \cdot \frac{I_m}{\sqrt{2}} = \frac{V_m I_m}{2}$.. Option D is often confused due to RMS values, but peak power involves peak values $V_m$ and $I_m$.

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

Exam Tip: Remember that while average power uses RMS values ($P = V_{rms}I_{rms}$), peak power uses amplitudes ($P_{peak} = V_m I_m$). Always check if the question asks for amplitude or RMS.

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