An LCR series circuit with `R = 100 Omega` is connected to a 300V, 50Hz a.c. source. If the capacitance is removed from the circuit then the current l

Question

An LCR series circuit with `R = 100 Omega` is connected to a 300V, 50Hz a.c. source. If the capacitance is removed from the circuit then the current l

An LCR series circuit with `R = 100 Omega` is connected to a 300V, 50Hz a.c. source. If the capacitance is removed from the circuit then the current lags behind the voltage by `30^(@)`. But if the inductance is removed form the circuit the current leads the voltage by `30^(@)`. What is the current in the circuit?
A. 2A
B. 3A
C. 1.5A
D. 4.5A

Monjur Alahi

4 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 - B
If C is removed, then the remaining circuit becomes an L-R circuit and the phase angle is given by
`tan phi = (X_L)/(R) = (omega L)/(R)`
Similarly if the capacitative reactance `X_(C) = 1/(omega C)`
Since both L and C are producing the same change in the phase angle, `:. omega L = 1/(omega C)`.
`:. I_(rms) = (E_(rms))/(R) = 300/100 = 3A`.