A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is I 0 . If the value of resistance R becomes twice of its initial value then amplitu

Question

A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is $I_{0}$ . If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be

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Answer:

C

Solution:

Initially, $I_{0} = \frac{ε _{m}}{R}$ Finally, $I_{0}^{1} = \frac{ε _{m}}{2 R} = \frac{I _{0}}{2}$

TestHub · Accepted Answer

Initially, I 0 ​ = R ε m ​ ​ Finally, I 0 1 ​ = 2 R ε m ​ ​ = 2 I 0 ​ ​

Physics - EMI/AC Question with Solution | TestHub

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