An LCR series circuit is under resonance. If `I_(m)` is current amplitude, `V_(m)` is voltage amplitude, R is the resistance, Z is the impedance, `X_(

Question

An LCR series circuit is under resonance. If `I_(m)` is current amplitude, `V_(m)` is voltage amplitude, R is the resistance, Z is the impedance, `X_(

An LCR series circuit is under resonance. If `I_(m)` is current amplitude, `V_(m)` is voltage amplitude, R is the resistance, Z is the impedance, `X_(L)` is the inductive reactance and `X_(C )` is the capacitive reactance, then
A. `I_(m)=(Z)/(V_(m))`
B. `I_(m)=(V_(m))/(X_(L))`
C. `I_(m)=(V_(m))/(X_(C ))`
D. `I_(m)=(V_(m))/(R )`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
Impedance of the circuit, `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))` At resonance, `X_(L)=X_(C )`
`therefore Z=R therefore I_(m)=(V_(m))/(Z)=(V_(m))/(R )`.

5 views Answered 2023-02-05 15:29