In the figure is shown a circuit with R=0.214? and L=0.227 mH. The AC generator

Question

In the figure is shown a circuit with R=0.214? and L=0.227 mH. The AC generator produces 25.3 ?V (rms) at 1160.0 kHz.

(a)

Find the capacitance C 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.

pF ( ± 0.2 pF)

(b)

For this value of C, find the rms voltage between points A and B. Hint: Find the impedance of the circuit. Then find the current.

mV ( ± 2 mV)

(c)

Repeat part (b), if the frequency of the generator is changed to1172.8 kHz without changing the values of V, R, L, and C. Warning: When the frequency changes, so does the impedance of the circuit. 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.

mV ( ± 0.04 mV)

Explanation / Answer

Here ,

a) for reactance of capacitor equal to reactance of inductor ,

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

frequency = 1/(2pi*sqrt(0.217 *10^-3 * C))

1160 *10^3 = 1/(2pi*sqrt(0.217 *10^-3 * C))

solving for C

C = 8.675 *10^-11 F

the capacitance is 8.675 *10^-11 F

b)

impedance of the ciruit , Z = R

Z = 0.214 ohm

I = 25.3/0.214 uA

I = 118.22 uA

Now, voltage across A and B , Vc = I*Xc

voltage across A and B , Vc = 118.22 *10^-6 *(1/(2pi*1160 *10^3 * 8.675 *10^-11))

voltage across A and B , Vc = 0.187 V

the voltage across A and B is 0.187 V

c)

Here , for change in frequency ,

Z = sqrt(0.214^2 + (2*pi*1172.8 *10^3 * 0.217 *10^-3 - 1/(2pi*1172.8 *10^3 * 8.675 *10^-11)))

Z = 5.89 Ohm

reactance of the capacitor ,

Xc = 1/(2pi*1172.8 *10^3 * 8.675 *10^-11)

Xc = 1564.3 ohm

Now, for votlage across A and B

Vc = 1564.3 * 25.3/5.89

Vc = 6719.31 uV

the voltage across A and B is 6719.31 uV

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