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If the three elements, L,C and R are arr...

If the three elements, `L,C` and `R` are arranged in parallel. Source has emf `230 V` and `L = 5.0 H, C = 80 mu F` and `R = 40 Omega`

A

The minimum impedance in the circuit is `40 Omega`

B

The maximum impedance in the circuit is `40 Omega`

C

The impedance is minimum at `omega = 50 "rads"^(-1)` of the source.

D

The impedance is maximum at `omega = 50 "rads"^(-1)` of the source

Text Solution

Verified by Experts

The correct Answer is:
2,3
Resonating angular frequency
`omega = (1)/(sqrt(LC)) = (1)/(sqrt(5 xx 80 xx 10^(-6))) = 50 "rads"^(-1)`
`:.` Resonance of `L` and `C` in parallel can be calculated.
`(1)/(X) = (1)/(X_(L)) - (1)/(X_(C )) = (1)/(omega L) - omega C`
Impedance of `R` and `X` in parallel is given by
`(1)/(Z) = sqrt((1)/(R^(2)) + (1)/(X^(2)))`
At resonating frequency of series `LCR, X_(L) = X_(C )`
So, `(1)/(X) = (1)/(X_(L)) - (1)/(X_(C )) = 0`
Thus, impedances `Z = R` and will be maximum, Hence, in parallel resonant circuit, current is minimum at resonant frequency.
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