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Physics · 2025 · 3 marks
Alternating CurrentRepresentation of AC Current and Voltage by Rotating Vectors — Phasors3 marksApplyshort answer
CBSE 2025 · Region 2 · Set 2 · Q24
Differentiate between the peak value and root mean square value of an alternating current. Derive the expression for the root mean square value of…
Differentiate between the peak value and root mean square value of an alternating current. Derive the expression for the root mean square value of alternating current, in terms of its peak value.
Marking-scheme solution
Peak value is the maximum value of the alternating current.
rms current is the equivalent dc current that would produce the same average power loss as the alternating current.
$\displaystyle I_{\mathrm{rms}}=\frac{I_{0}}{\sqrt{2}}$
$\displaystyle I_{\mathrm{rms}}=0.707 I_{0}$
The instantaneous power dissipated in the resistor is $\displaystyle P=i^{2} R=i_{m}^{2} R \sin ^{2} \omega t$
The average power over a cycle is:
$\displaystyle \bar{P}=\left\langle i^{2} R\right\rangle=\left\langle i_{m}^{2} R \sin ^{2} \omega t\right\rangle$
$\displaystyle \left\langle\sin ^{2} \omega t\right\rangle=\frac{1}{2}$
$\displaystyle \bar{P}=\frac{1}{2} i_{m}^{2} R=I_{\mathrm{rms}}^{2} R$
$\displaystyle I_{\mathrm{rms}}=\sqrt{\frac{i_{m}^{2}}{2}}=\frac{i_{m}}{\sqrt{2}}$
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CBSE Class 12 Physics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.