JEE Mains · Physics · STD 12 -7. Alternating current
An \(AC\) current is given by \(I = I _{1} \sin \omega t + I _{2} \cos \omega t\). A hot wire ammeter will give a reading
- A \(\sqrt{\frac{I_{1}^{2}-I_{2}^{2}}{2}}\)
- B \(\sqrt{\frac{ I _{1}^{2}+ I _{2}^{2}}{2}}\)
- C \(\frac{ I _{1}+ I _{2}}{\sqrt{2}}\)
- D \(\frac{ I _{1}+ I _{2}}{2 \sqrt{2}}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{ I _{1}^{2}+ I _{2}^{2}}{2}}\)
Step-by-step Solution
Detailed explanation
\(I=I_{1} \sin \omega t+I_{2} \cos \omega t\) \(\therefore I_{0}=\sqrt{I_{1}^{2}+I_{2}^{2}}\) \(\therefore I_{ rms }=\frac{I_{0}}{\sqrt{2}}=\sqrt{\frac{I_{1}^{2}+I_{2}^{2}}{2}}\)
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