Question 8: In the eireuit showa on the riht, R 30n L-Imt, and C80pF. A sinusoid

Question

Question 8: In the eireuit showa on the riht, R 30n L-Imt, and C80pF. A sinusoidal voltage u with a t.ass, amplitude of 100mV is applied. a) At what frequeney will the voltage acroes R be largest? Cal culate the size of this voltage, and the power dissipated by R at this frequency b) If C is increased to 120 pF, and the frequency remsins the same as in part a, what will be the new voltage across R? What power will be dissipated by R2 e) Based on your results from a and b, can you explain how one might construct a slimple radio tuner?

Explanation / Answer

a) at resonant frequency voltage across R becomes larget.

resonant frequency, f = 1/(2*pi*sqrt(L*C))

= 1/(2*pi*sqrt(1*10^-3*80*10^-12))

= 5.63*10^5 Hz

at resonant frequency,
impedance of the circuit, z = R

= 30 ohms

voltage across the resistor = 100 mV

power dissipated by R, P = I^2*R

= (V/R)^2*R

= (0.1/30)^2*30

= 3.33*10^-4 W

b) inductive reactance, XL = 2*pi*f*L

= 2*pi*5.63*10^5*1*10^-3

= 3537 ohm

capacitive reactance, Xc = 1/(2*pi*f*C)

= 1/(2*pi*5.63*10^5*120*10^-12)

= 2356 ohm

impedance of the ciruit, z = sqrt(R^2 + (XL - Xc)^2)

= sqrt(30^2 + (3537 - 2356)^2)

= 1181 ohm

I = V/z

= 0.1/1181

= 8.47*10^-5 A

P = I^2*R

= (8.47*10^-5)^2*30

= 2.15*10^-7 W

c) at resonant frequency only current in the circuit becomes maximim.

so, a radio tuner always tuned to resonant frequency.

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