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GUJCET · Physics · Alternating Current
An electric current has both DC and AC components. DC component of 8 A and AC component is given as \(I=6 \sin \omega t\). So rms value of resultant current is ____________
- A 8.05 A
- B 9.05 A
- C 11.58 A
- D 13.58 A
Answer & Solution
Correct Answer
(B) 9.05 A
Step-by-step Solution
Detailed explanation
(B) 9.05 A
\(\langle\mathrm{I}\rangle=\langle 8+6 \sin \omega\rangle\)
\(\therefore\left\langle\mathrm{I}^{2}\right\rangle=\left\langle(8+6 \sin \omega t)^{2}\right\rangle\)
\(\therefore\left\langle\mathrm{I}^{2}\right\rangle=\langle 64+96 \sin \omega t\)
\(
+36 \sin ^{2} \omega t
\)
\(\therefore\left\langle\mathrm{I}^{2}\right\rangle=64+96(0)+36\left(\frac{1}{2}\right)\)
\(\left\langle\mathrm{I}^{2}\right\rangle=64+18\)
\(\mathrm{I}_{r m s}=\sqrt{\left\langle\mathrm{I}^{2}\right\rangle}\)
\(\mathrm{I}_{r m s}=\sqrt{82}\)
\(\mathrm{I}_{r m s}=9.05 \mathrm{~A}\)
\(\langle\mathrm{I}\rangle=\langle 8+6 \sin \omega\rangle\)
\(\therefore\left\langle\mathrm{I}^{2}\right\rangle=\left\langle(8+6 \sin \omega t)^{2}\right\rangle\)
\(\therefore\left\langle\mathrm{I}^{2}\right\rangle=\langle 64+96 \sin \omega t\)
\(
+36 \sin ^{2} \omega t
\)
\(\therefore\left\langle\mathrm{I}^{2}\right\rangle=64+96(0)+36\left(\frac{1}{2}\right)\)
\(\left\langle\mathrm{I}^{2}\right\rangle=64+18\)
\(\mathrm{I}_{r m s}=\sqrt{\left\langle\mathrm{I}^{2}\right\rangle}\)
\(\mathrm{I}_{r m s}=\sqrt{82}\)
\(\mathrm{I}_{r m s}=9.05 \mathrm{~A}\)
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