In a Certain Series LCR Circuit

By BYJU'S Exam Prep

Updated on: September 25th, 2023

A certain series L, C, R circuit with L=10 mH and capacitance of L=10 μF when connected to an alternating source of voltage 100 V draws a maximum current of 7 amp (rms value) when the frequency of the source is adjusted for the maximum power condition. What is the value of approximate resistance of the circuit?

1. 10.22Ω
2. 21.22Ω
3. 7.22Ω
4. 5.12Ω

In the certain series, LCR circuit resistance R=10.22.

Solution

Power maximum power to occur in series LCR circuit the system should be under resonance,

That means angular frequency ω=1/√LC

So, at ω=1/√LC series LCR circuit will under resonance, and at resonance maximum power will be dissipated in resistor.

L=10 mH×√10×103=10√10

1C=1/10 μF×√10×103=√10×10

Then impedance Z=√R2+(XLXC)2

At resonance  XL=XC

So, impedance Z=R

Maximum current Imax=7√2 A (in LCR circuit maximum current will occur at resonance)

Imax=Vmax/Z

Imax=Vmax/R

7√2 A=100 V/R

R=100 V/7√2 A=10.101 ohm.

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