Consider an L-R-C series cirquit composed of the inductor, resistor, and capacit

Question

Consider an L-R-C series cirquit composed of the inductor, resistor, and capacitor where you have a 200 ohm resistor, a .400-H inductor, and a 6.00-uF capacitor. The circuit is connected to an ac source with voltage amplitude 30.0 V
a) at what frequency (in Hz) is the circuit in resonance?
b) sketch the phasor diagram at the resonance frequency (I, V_R,V_L, V_C, V_total) ****V_R is the same as V subscript R****
c) what are the voltage amplitudes across the resisotr, inductor, and capacitor when the source frequency equals the resonance frequency?
d) what is the average power for the circuit?

Explanation / Answer

a) RLC circuit is in resonance when the impedance due to capacitor and inductor is the same. This gives the resonance frequency as = 1/(LC) rad/s.

so resonance frequency is 1/(0.4*6*10-6)= 645.536 rad/s = 102.74 Hz

b) As the impedances due to capacitor and inductor are equal and opposite the circuit just sees a real resistance, and hence the current is in phase with the voltage and keeps rotating with an angular speed equal to the resonance frequency. Same is the case for V_R as Current which is a real quantity here (due to cancellation of impedances) is multiplied by another real quantity R. For V_L and V_C they will be 900 out of phase with current because of multiplication with impedance terms. V_L will be 900 ahead of I and V_c 900 behind it. V_Total again will be in phase with I rotating along with it.

c) To get this we need the amplitude of current through the circuit which can simply be found as, I = V/R (as the impedances cancel out each other at resonance).

I = 30/200 = 0.15 amp.

V_R = I*R = 30 V

V_L = I*Xc = I/(C) = 0.15/(645.536*6*10-6)= 38.73

V_C = 38.73 (1800 out of phase with V_L cancelling it)

d) Average power is found as IRMS2*R = 0.15/2*0.15/2*200 = 2.25 Watt

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