22.13 Part A Compute the impedance of a series R-L-C circuit at angular frequenc

Question

22.13

Part A

Compute the impedance of a series R-L-C circuit at angular frequencies of ?1= 1000rad/s , ?2= 765rad/s and ?3= 485rad/s . Take R = 220? , L = 0.865H and C = 2.00?F .

Enter your answers as three numbers separated with commas.

Z1, Z2, Z3 = ________ ?

Part B

Describe how the current amplitude varies as the angular frequency of the source is slowly reduced from 1000rad/s to 485rad/s .

Part C

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

O Amplitude is always constant. O Amplitude decreases. O Amplitude increases. O First amplitude increases next it decreases. O First amplitude decreases next it increases. 22.13 Part A Compute the impedance of a series R-L-C circuit at angular frequencies of ?1= 1000rad/s , ?2= 765rad/s and ?3= 485rad/s . Take R = 220? , L = 0.865H and C = 2.00?F . Enter your answers as three numbers separated with commas. Z1, Z2, Z3 = ________ ? Part B Describe how the current amplitude varies as the angular frequency of the source is slowly reduced from 1000rad/s to 485rad/s . Part C What is the phase angle of the source voltage with respect to the current when ? = 1000 rad/s?

Explanation / Answer

R = 220 ohm

L = 0.865 H

C = 2*10^-6 F

A)

Impedance is given by :

Z = sqrt(R^2+(W*L - 1/(W*C))^2)

where W = angular frequency

So, for W1 = 1000 rad/s

Z1 = sqrt(220^2+(1000*0.865 - 1/(1000*2*10^-6))^2) = 426.2 ohm

for W2 = 765 rad/s

Z2 = sqrt(220^2+(765*0.865 - 1/(765*2*10^-6))^2) = 220.2 ohm

for W3 = 485 rad/s

Z3 = sqrt(220^2+(485*0.865 - 1/(485*2*10^-6))^2) = 649.8 ohm

B)

We know, current, I = V/Z

As W is decreased from 1000 rad/s to 485 rad/s, the impedance Z first decreases from 426.2 ohm to 220.2 ohm and then increases to 649.8 ohm.

As current is inversely proportional to Z , so current amplitude will first increase and then decrease <---------answer

c)

phase angle, = acos(R/Z) = acos(220/(426.2)) = 58.9 degrees <-------answer

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