9-3. In the figure is shown a circuit with R = 0.214 and L = [032,233 mh. The AC

Question

9-3. In the figure is shown a circuit with R = 0.214 and L = [032,233 mh. The AC generator produces 253 V (rms) at 1160.0 kHz. (a) Find the capacitance C (pF) such that the magnitude of its reactance is equal to the magnitude of the inductor's reactance. This brings the circuit into resonance with the frequency of the generator. (b) For this value of C, find the rms voltage (m V) between points A and B. Hint: Find the impedance of the circuit. Then find the current. (e) Repeat part (b) f the frequency of he current. the generator is changed to [04] and C. Warning: When the frequency changes, so does the impedance of the circuit. Hint: Calculate Xc and XL to 5 significant figures. You also need the value of C from part (a) to 5 significant figures. Note: This is how a radio tunes into a station. The generator represents an antenna which picks up an AC voltage from radio waves (for example, KSL, which broadcasts at 1160 kHz). When you tune the radio, you are adjusting a capacitor so that an LC circuit is in resonance with the frequency of the wave. This causes the signal to be amplified across the capacitor. Other signals with nearby frequencies are not amplified, so you hear only the one radio station. kHz without changing the values of V, R, L,

Explanation / Answer

a)

Given XL=XC

2pifL =1/2pifC

=>C=1/4pi2f2L =1/4pi2*(1160*103)2*0.233*10-3

C=8.08*10-11F

b)

Maximum Current

I=(25.3*10-6)/0.214=1.1822*10-4A

Voltage across capacitor is

VC =(1.1822*10-4)(1/2pi*1160*103*8.08*10-11)

VC=0.2 Volts

c)

Inductive reactance

XL=2pifL =2pi*(1170.1*103)(0.233*10-3)=1713 ohms

Capacitive reactance

XC =1/2pifC =1/2pi*(1170.1*103)(8.08*10-11)=1683.4 ohms

Impedance

Z=sqrt(R2+(XL-XC)2)=sqrt(0.2142+(1713-1683.4)2)=29.6 ohms

Total current

I=25.3*10-6/29.6=8.547*10-7 A

Voltage across capacitor is

VC =(8.547*10-7)*1683.4

VC=1.4388*10-3 Volts

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