KCET · Physics · Alternating Current
An alternating current is given by \(i=i_1 \sin \omega t+i_2 \cos \omega t\). The rms current is given by
- A \(\frac{i_1-i_2}{\sqrt{2}}\)
- B \(\sqrt{\frac{i_1^2+i_2^2}{2}}\)
- C \(\sqrt{\frac{i_1^2+i_2^2}{\sqrt{2}}}\)
- D \(\frac{i_1+i_2}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{i_1^2+i_2^2}{2}}\)
Step-by-step Solution
Detailed explanation
Equation of alternating current,
\(i=i_1 \sin \omega t+i_2 \cos \omega t \)
\( \text {Let } i_1=I \cos \phi \)
\( \text {and } i=I \sin \phi \)
\( i=I \cos \phi \sin \omega t+I \sin \phi \cos \omega t \)
\( =I(\sin \omega t \cos \phi+\cos \omega t \sin \phi) \)
\( \Rightarrow i=I \sin (\omega t+\phi) \)
\( \therefore \text {rms value of } i_1 \text {, } \)
\( i_{\mathrm{rms}}=\frac{I}{\sqrt{2}}\)
From Eqs. (i) and (ii), we get
\(i_1^2+i_2^2=I^2 \cos ^2 \phi+I^2 \)
\(=I^2 \)
\(\Rightarrow \sqrt{i_1^2+i_2^2}=I\)
\(i_1^2+i_2^2=I^2 \cos ^2 \phi+I^2 \sin ^2 \phi=I^2\left(\cos ^2 \phi+\sin ^2 \phi\right)\)
From Eqs. (iii) and (iv), we get
\(i_{\mathrm{rms}}=\sqrt{\frac{i_1^2+i_2^2}{2}}\)
\(i=i_1 \sin \omega t+i_2 \cos \omega t \)
\( \text {Let } i_1=I \cos \phi \)
\( \text {and } i=I \sin \phi \)
\( i=I \cos \phi \sin \omega t+I \sin \phi \cos \omega t \)
\( =I(\sin \omega t \cos \phi+\cos \omega t \sin \phi) \)
\( \Rightarrow i=I \sin (\omega t+\phi) \)
\( \therefore \text {rms value of } i_1 \text {, } \)
\( i_{\mathrm{rms}}=\frac{I}{\sqrt{2}}\)
From Eqs. (i) and (ii), we get
\(i_1^2+i_2^2=I^2 \cos ^2 \phi+I^2 \)
\(=I^2 \)
\(\Rightarrow \sqrt{i_1^2+i_2^2}=I\)
\(i_1^2+i_2^2=I^2 \cos ^2 \phi+I^2 \sin ^2 \phi=I^2\left(\cos ^2 \phi+\sin ^2 \phi\right)\)
From Eqs. (iii) and (iv), we get
\(i_{\mathrm{rms}}=\sqrt{\frac{i_1^2+i_2^2}{2}}\)
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