Question

The provided series RLC circuit is in a sinusoidal steady state at a frequency of 60 Hz. V = 100 V, R = 20 L = 15 mH.C = 150 F (t) Cl2 a) Calculate the magnitude (12) and phase angle () of the load. XL2WL عليا function of z = f(R, XL, XC) M = f(R, XL,XC) b) Calculate the source currentl. JE VLO I ZLO c) Calculate the instantaneous source current it). i(+) = S2 I cos (wt-)

I need help completing the answer the formula's is given for each part. Thank you.

Answer 1

Question (1):

Following values are given in the question:

V = 100V

f = 60 Hz

R = 2 ohm

L = 15 mH = 15*10-3 H

C = 150 $\mu$ F = 150*10-6 F

(a)

So, Xl = $\omega$ L = (2$\pi$f)L

Xl = 2*$\pi$*60*15*10-3

Xl = 5.65 ohm

Xc = 1/$\omega$C = 1/(2$\pi$f)C

Xc = 1/(2*$\pi$*60*150*10-6)

Xc = 17.68 ohm

So,   

12 = R2 + (X. - X1)

|Z| = $\sqrt{}$ [22 + (17.68-5.65)2]

|Z| = $\sqrt{}$ 148.72

|Z| = 12.19 ohm

Phase angle ($\phi$) =

(XC-X1) arctan R

$\phi$ =

(17.68 – 5.65) arctan 2

$\phi$ =

arctan 6.01

$\phi$ = 80.56 degrees

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(b)

If we consider voltage as refrence, then voltage phasor V = 100$\angle$0

So, source current I = V/ Z$\angle$$\phi$

I = 100$\angle$0/12.19$\angle$80.56

I = 8.2$\angle$-80.56 A

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(c)

Instantaneous current i(t) = $\sqrt{}$ 2 Icos($\omega$t - $\phi$ )

Substituting the values in the above expression we get

i(t) = $\sqrt{}$ (2)*8.2 cos[(2*$\pi$*60)t - 80.56]

i(t) = 11.59 cos(377t - 80.56) A

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Question (2):

Given in the question

V = 120$\angle$0

Xl = j0.2 ohm

Xc = -j3 ohm

R = 5 ohm

(a)

Circuit impedance Z = Xl+ (R*Xc)/(R+Xc)

Z = j0.2 + [5*(-j3)]/[5-j3]

Z = j0.2 - [-1.32 + j2.2]

Z = 1.32 - j2

Z = 2.39$\angle$-56.57 ohm

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(b)

Current I = V/Z

I = 120$\angle$0/2.39$\angle$-56.57

I = 50.2$\angle$56.57 A

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(c)

Active power P = VIcos$\phi$ where $\phi$ is the phase angle of the load equal to -56.57 degrees

P = 120*50.2cos(-56.57)

P = 6024*0.55

P = 3318.72 W

P = 3.318 KW

Reactive power Q = VIsin$\phi$

Q = 120*50.2sin(-56.57)

Q = 6024*(-0.83)

Q = -5027.38 VAR

Q = -5.027 KVAR

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(d)

phase angle $\phi$ = -56.57 degrees

So, power factor cos$\phi$ = cos(-56.57) = 0.55