Pre lab 4 Result sheet
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ECE209 – Fundamentals of Electrical Engineering Lab 4: Phasors and AC Power
Lab 4 – Pre lab: Phasors and AC Power
ECE209: Fundamental of Electrical Engineering
Name
Student ID
CCID
Harshith rao
1760726
hhrao
Phasors and AC Power - (RCL)
Resistor -200Ω
Capacitor - 1
μ
F
Inductor - 2.5mH
Frequency
Hz
4.0 k
796
12.7 k
Z
Ω
200 200.1
199.7 a.
Calculate the impedance at the applied frequency for the
components in the Phasors and AC Power section of the lab 4 manual. Show your
work and place your results in the appropriate place in this results sheet. (You can
neglect R
SENSE
for calculation.)
For inductor= L= 2.5mH
XL= wL 2pi*f*L= 2pi*12.7*1000*2.5*10-3 = 199.49 ohms Z= sqrt(XL^2+ R^2) = sqrt( 199.49^2+10^2)= 199.7 For capacitor 1 uF = Xc= 1/2pifc 1/2pi*796*1*10^-6 = 199.94ohms Z= sqrt( r^2+XC^2) = sqrt( 10^2+199.94^2)= 200.1
For resistor 200 ohms Z= R+0j= 200 Page 1 of 4
ECE209 – Fundamentals of Electrical Engineering Lab 4: Phasors and AC Power
Series RLC Circuit - (KVL)
f
s
kHz
1.80
3.20
4.60
Z
Ω
116.7
100.00
106.9
b.
Calculate the total impedance of the Series RLC circuit
(figure 7 in lab 4 manual)
for 3 different applied frequencies. Show your work for
the 1.8 kHz case. Place your results in the appropriate place in this result sheet.
For 1.8khz case Total impedance formula Z= sqrt(R^2+(XL-XC) ^2
XL= wL= 2pi*1.8*1000*2.5*10^-3 XL= 28.27 ohms XC= 1/2pi*f*C= 1/2pi*1.8*1000*(1*10^-6) = 88.42ohms Z= sqrt(R^2+(XL-XC) ^2)
= sqrt( (100)^2+(28.27-88.42)^2)) = 116.7 ohms Page 2 of 4
ECE209 – Fundamentals of Electrical Engineering Lab 4: Phasors and AC Power
Parallel RC Circuit - (KCL)
f
s
Hz
400
800
1.20 k
Z
Ω
178.7
141.05
110.53
c.
Calculate the total impedance of the Parallel RC circuit
(figure 9 in lab 4 manual) for 3 different applied frequency. Show your work for
the 400 Hz case. Place your results in the appropriate place in this result sheet.
(You can neglect the value of
R
SC
and R
SS while calculating impedance)
For 400 ohms Impedance formula for a parallel RC circuit= Z= 1/sqrt((1/R)^2+(1/XC)^2 Xc= 1/2pi*f*C = 1/2pi*400*1*10^-6= 398ohms R= 100+100= 200 R= 200 ohms Z= 1/sqrt(1/200)^2+(1/398)^2)
Z= 178.7 ohms d.
Describe in your own words what the Power Triangle is. Use an example to demonstrate.
A power triangle is basically a way of representing the power consumed in an AC
circuit by 3 components which are impedance, reactance, and resistance.
Impedance is not the sum of reactance and resistance in a circuit as both reactance
and resistance are out of phase with each other by 90 degrees. Because they are out
of phase with each other by 90 we can create a right-angled triangle where
Page 3 of 4
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ECE209 – Fundamentals of Electrical Engineering Lab 4: Phasors and AC Power
resistance R and reactance X are at 90 with each other and impedance Z is the
hypotenuse Page 4 of 4
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