AP EAMCET · PHYSICS · Alternating Current
An alternating current is given by \(\mathrm{i}=(3 \sin \omega t+4 \cos \omega t)\)
A. The \(r m s\) current will be
- A \(\frac{7}{\sqrt{2}} A\)
- B \(\frac{1}{\sqrt{2}} A\)
- C \(\frac{5}{\sqrt{2}} A\)
- D \(\frac{3}{\sqrt{2}} A\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{\sqrt{2}} A\)
Step-by-step Solution
Detailed explanation
\(i=(3 \sin w t+4 \cos w k) A\) \(\therefore\) Peak current, \(\mathrm{I}_0=\sqrt{3^2+4^2}=5 \mathrm{~A}\) \(\therefore \quad\) RMS current, \(\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{\mathrm{o}}}{\sqrt{2}}=\frac{5}{\sqrt{2}} \mathrm{~A}\)
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