Physics Alternating Current ETEA 2015
PMDC Verified Question 106 of 127
Peak value of alternating current is \(5\sqrt{2}\text{ A}\). The mean square value of current will be:
A
5 A
B
\(25\text{ A}^2\)
C
\(5\sqrt{2}\text{ A}\)
D
\(50\text{ A}^2\)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \(25\text{ A}^2\)
Concept:

The mean square value of an alternating current waveform is half of the square of its peak value.

Formula / Reaction:

$$\langle I^2 \rangle = \frac{I_0^2}{2}$$

Solution:

  • Given peak current: \(I_0 = 5\sqrt{2}\text{ A}\).


  • \(I_0^2 = (5\sqrt{2})^2 = 25 \times 2 = 50\text{ A}^2\).


  • \(\langle I^2 \rangle = \frac{50}{2} = 25\text{ A}^2\).


Why other options are incorrect:

  • Opt_A: \(5\text{ A}\) is the RMS value (\(I_{\text{rms}} = \sqrt{25} = 5\text{ A}\)), not the mean square value.


  • Opt_C: \(5\sqrt{2}\text{ A}\) is the peak current \(I_0\).


  • Opt_D: \(50\text{ A}^2\) is the peak square \(I_0^2\), without dividing by 2.

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