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AN GRG cIRcuiT i5 ARwe 3Y A TIME VARYING VOLTAGE
Su PPLY WITH A VOLTA QG Of N Lt) = 100V.cos (ut+4).
Aw LRC CIRCUIT IS PRIVEN BY A TIME VARYIN be VOLTAGG
FindS
Wees
1f La ZmlH, C=Amf, R=2e
FIND ALL VALUES FOe waWRO:
%3D
工
R
Transcribed Image Text:AN GRG cIRcuiT i5 ARwe 3Y A TIME VARYING VOLTAGE Su PPLY WITH A VOLTA QG Of N Lt) = 100V.cos (ut+4). Aw LRC CIRCUIT IS PRIVEN BY A TIME VARYIN be VOLTAGG FindS Wees 1f La ZmlH, C=Amf, R=2e FIND ALL VALUES FOe waWRO: %3D 工 R
Expert Solution
Step 1

Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered.

A Series LCR Circuit

Advanced Physics homework question answer, step 1, image 1

The figure above shows a series LCR circuit. The input voltage is given by V=V0sinωt+ϕ where V0 is the amplitude of the voltage. 

The current amplitude is given by

I0=V0R2+ωL-1ωC2

The denominator is known as the impedance of the circuit. At resonance, the inductive reactance becomes equal to the capacitive reactance

ωresL=1ωresCωres=1LC

This is the resonant frequency. At resonance, the impedance becomes only resistive and hence the peak current at resonance becomes

I0=V0R

The potential difference across the resistor is

VR=I0R=V0

The potential difference across the inductor

VL=ωresLI0

The potential difference across the capacitor

VC=I0ω0C

Step 2

In the given question the resistance is R=2 Ω

        the inductance is L=2 mH=2×10-3 H

        the capacitance is C=0.5 mF=0.5×10-3 F

        the input voltage V=100 cosωt+ϕ V

a) The resonant frequency is given by

ωres=1LC=12×10-3×0.5×10-3=1000 rad/s

 

b) The peak value of current flowing in the circuit at resonance

I0=V0R=1002A=50 A

The peak value of the potential difference across the resistor is VR=V0=100 V.

 

c) The peak value of potential difference across the inductor is 

VL=ωres LI0=1000×2×10-3×50 V=100 V

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