Question

(a) Compute the impedance of a series R-L-C circuit at angular frequencies of w1 = 1000 rad/s, w2 = 735 rad/s and w3 = 540 rad/s. Take =160 ohm, = 0.940 H and = 2.05 uF.

(b) Describe how the current amplitude varies as the angular frequency of the source is slowly reduced from 1000 to 540 .
a-Amplitude is always constant.
b-Amplitude decreases.
c-Amplitude increases.
d-First amplitude increases next it decreases.
e-First amplitude decreases next it increases.

(c) What is the phase angle of the source voltage with respect to the current when w = 1000 rad/s?

Answer 1

a)

in general

for w = 1000 rad/s

xL = 1000 * 0.94 = 940

xC = 1/1000*2.05 uF = 487.81

Z1 = sqrt(160^2 + (940-487.81)^2) = 479.66 ohms

for w = 735 rad/s

xL = 735 * 0.94 = 690.9

xC = 1/735*2.05 uF = 663.68

Z2 = sqrt(160^2 + (690.9-663.68)^2) = 162.30 ohms

for w = 540 rad/s

xL = 540 * 0.94 = 507.6

xC = 1/540*2.05 uF = 903.34

Z3 = sqrt(160^2 + (507.6-903.34)^2) = 426.86 ohms

b) The amplitude of the current will be given by:

, assuming the amplitude of the voltage source is a constant, we see current will increase as impedance is reduced, and current will decrease as impance is increased.

From Z1, the current is at some value, V/Z1 then it increases since Z2 < Z1, then it decreases again as Z3 > Z2.

Choice (d) is the answer. This is expected as the resonant frequency is: when wo = 1/sqrt(LC) =720.38 , at the resonent frequency, the impedance is minimum(it's purely resistive), so the current is maximum.

c)

The phase angle of the source(at ?=1000 rad/s) will the same phase angle of the impedance, and the phase angle is

arccos(R/Z) = arccos(160/479.66) = 70.51 degrees

The phase angle could also be obtained by:

In this case, they will all be the same, but if the capacitive reactance was bigger than the inductive reactance, then the arcsin and arctangent would give you negative angles which implies that the current is leading the voltage. For your info, you should always use the arcsin or arctangent, because they take the sign of the angle into account. Fro cosine though, you will always get a positive angle. the cosine is only right when the inductive reactance is bigger than the capacitive reactance.