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For the circuit of figure Ex32.33,

a. What is the resonance frequency, in both rad/s and Hz?

b. Find VrandVcat resonance.

c. How canVcbe larger thanε0? Explain .

Short Answer

Expert verified

(a) Resonance frequency in rad/s=1010rad/s

Resonance frequency in Hz=49.672Hz

(b)At resonance,Vr=10V,Vc=10410V

(c) See explanation below .

Step by step solution

01

Part(a) Step1: Given information.

We have given,

L=1mHC=1μF

We have to find resonance frequency in rad/s(ω) and Hz.

02

Part(a) Step2: Simplify.

We know at resonance condition ,the frequency is;

ω=1LCrad/s

ω=110-3×10-6rad/s (Substitutingthevalues)

ω=1010rad/s

We know ,

ω=2πf

f=ω2π

f=10102π (Substitutingthevalues)

f=49.672Hz

03

Part(b) Step1: Given information.

We have given,

R=10ohmC=1μFε0=10Vω=1010rad/s

We have to find Vrand Vc.

04

Part(b) Step2: Simplify.

Firstly we need to find Xc ,

Xc=1ωC

Xc=11010×10-6ohm (Substituingthevalues)

Xc=104×10ohm

At resonance, I is maximum and equal to ε0R,

I=ε0R

I=10V10ohm (Substitutingthevalues)

I=1A

now,

Vr=IR

Vr=1×10V (Substitutingthevalues)

Vr=10V

Also,

Vc=IXc

Vc=1×10410V (Substitutingthevalues)

Vc=10410V

05

Part(c) Step1: Given information.

We have givenVCandε0.

We have to explain how canVcbe larger thanε0.

06

Part(c) Step2: Explanation.

If the circuit is in resonance condition ω=1LC, the peak to peak voltages on capacitor and the inductor can be larger than the peak to peak voltage ε0.

Also'

Vc=XcZ×ε0 ( Z=totalimpedence)

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