In a series `LCR` circuit with an AC source, `R = 300 Omega, C = 20 muF, L = 1.0 henry, epsilon_(rms) = 50 V` and `v = 50/(pi) Hz`. Find (a) the rms c

Question

In a series `LCR` circuit with an AC source, `R = 300 Omega, C = 20 muF, L = 1.0 henry, epsilon_(rms) = 50 V` and `v = 50/(pi) Hz`. Find (a) the rms c

In a series `LCR` circuit with an AC source, `R = 300 Omega, C = 20 muF, L = 1.0 henry, epsilon_(rms) = 50 V` and `v = 50/(pi) Hz`. Find (a) the rms current in the circuit and (b) the rms potential differences across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.

Monjur Alahi

7 views

2025-01-04 04:42

1 Answers

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

Correct Answer - (a)`0.1 A` ,(b)`50 V, 30 V,10 V`(Note that the sum of the `rms` potential differences across the three elements is greater then the `rms` voltage of the source.)
(a)`I_(rms)=E_(rms)/z` where `z=sqrt(R^(2)+(omegaL-1/omegaC)^(2))`
(b)`V_(R rms)=I_(rms)R=30,V_(L rms)=I_(rms)(omegaL)=10`
`V_(C rms)=I_(rms)(1/(omegaC))=50`