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Alternating Current — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsAlternating Current1 Mark
MCQ
An alternating current is given by $\text{i}=\text{i}_1\cos\omega\text{t}+\text{i}_2\sin\omega\text{t}.$ The rms current is given by:
  • A
    $\frac{\text{i}_1+\text{i}_2}{\sqrt{2}}$
  • B
    $\frac{|\text{i}_1+\text{i}_2|}{\sqrt{2}}$
  • ✓
    $\sqrt{\frac{\text{i}_1^2+\text{i}_2^2}{2}}$
  • D
    $\sqrt{\frac{\text{i}_1^2+\text{i}_2^2}{\sqrt{2}}}$
✓

Answer

Correct option: C.
$\sqrt{\frac{\text{i}_1^2+\text{i}_2^2}{2}}$
$\text{i}=\text{i}_1\cos\omega\text{t}+\text{i}_2\sin\omega\text{t}$
$\text{I}_\text{rms}=\frac{\int\limits_0^\text{T}\text{I}^2\text{dt}}{\int\limits_0^\text{T}\text{dt}}$
if $\text{I}=\cos\omega\text{t}$
$\text{I}_\text{rms}^2=\frac{\text{I}_0^2}{2}$
$\text{i}=\text{i}_1\cos\omega\text{t}+\text{i}_2\sin\omega\text{t}$
Than $\text{i}_\text{rms}^2=\frac{\text{i}_1^2}{2}+\frac{\text{i}_2^2}{2}$
$\text{i}_\text{rms}=\sqrt{\frac{\text{i}_1^2+\text{i}_2^2}{2}}$

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