Vadodara Institute of Engineering
kotanbi-391510
Active learning Assignment
on
Single phase AC CIRCUIT
SUBMITTED BY:
1) Bhatiya gaurang.(13ELEE558)
2) Ghori jasminn.(13ELEE547)
3)Chorasiya monukumar.(13ELEE544)
4) Bhatt diveysh.(13ELEE548)
GUIDED BY:
BRANCH:ELECTRICAL-3
SUDHIR PANDAY.
YEAR:2013-2014
AC Circuits
• AC Current
•
peak-to-peak and rms
• Capacitive Reactiance
•
magnitude and phase
• Inductive Reactance
•
magnitude and phase
AC Circuits
AC Signals and rms Values
• There are a number of ways to describe the current or
voltage for a time varying signal: peak value, peak-topeak, and rms. Write in your own words what each means
and indicate the three values on the sine wave signal
shown below.
Vp
Vpp
Vrms
AC Circuits
• What is V average, Vave?
Vave  0
• How is Vrms related to Vm?
Vm 
2Vrms
We will derive this in a minute.
When you measure an ac current or voltage with a
DMM you are measuring rms values.
AC Circuits
• Connect the cables from the Signal/Function Generator and the voltage
input cables to the 2.2 kW resistor on the E&M board. The red and
black cables with the red and black plugs are the input to the voltage
sensor. Make sure the two grounds (black cables) are connected to the
same side of the resistor. Start DataStudio and configure the program
to measure voltage using channel A. You will recall that you do this
by clicking and dragging the plug icon to channel A and choosing
Voltage Sensor. For output we will connect both the digital output and
the oscilloscope. Connect both to channel A.
• Next select the Signal/Function Generator by clicking the Signal icon
on the lower left side of the DataStudio window. Set the amplitude to
5 volt and the frequency to 100 Hz on the function generator. Select
the sine wave (AC) and click ON.
• Measure the peak value and the peak-to-peak value using the
oscilloscope and measure the rms value using the digital meter.
AC Circuits
• Measure the peak value and the peak-to-peak value using
the oscilloscope and measure the rms value using the
digital meter.
Vm =
Vpp =
Vrms =
• What is the ratio of Vrms to Vp?
Vrms/Vp = 0.707
• Is this what you expect? Explain.
Yes, at least we expect the value to be less than one.
AC Circuits
• The rms value is found by squaring the signal, integrating
over one period, and then taking the square root. Lets do
it. First square the following signal.
V(t) = Vpsin(wt)
V2(t) = Vp2sin2(wt)
• Now integrate this with respect to time from t = 0 to t = T
(the period). Look in Appendix A page A-10 for the
integral.
t sin( 2wt ) 

 Vp sin (wt )dt Vp  sin (wt )dt  Vp  
0
0
4w  0
2
T
2
2
Hint: What is wT?
2
T
T
2
2
T sin( 2wT)  2 T
 V  
 Vp
2
4w 
2
2
p
• Finally, dividing by T and take
the square root you should get
Vrms 
Vp
2
AC Circuits
Resistance, Capacitance, and Inductance
• When we discuss resistors, capacitors, and inductors in a
circuit there are two important points to remember.
– The magnitude relationship between the current in
and voltage across a resistor, capacitor, or inductor.
– The phase relationship between the current in and
voltage across a resistor, capacitor, or inductor.
AC Circuits
Resistance
• The simplest case is a resistor in a circuit.
– The current and voltage are in phase
– The magnitude is VR=IR
The current and voltage are in phase in a resistor.
Current and voltage reaches a minimum at time T
AC Circuits
Capacitive Reactance
• When we discuss capacitors in a circuit there are two
important points for a capacitor. One is the phase
relationship between the current in and voltage across a
capacitor. What is the phase relationship? This is the
bold statement on page 855 in your text.
The current in a capacitor leads the voltage by 900.
Current reaches a minimum at time T
Voltage reaches a minimum at a later time T+DT
AC Circuits
• The second point is the magnitude relationship between
current and voltage for a capacitor. What is this
relationship?
Like
VC, p  I p XC
or
VC ,rms  IrmsXC
VR, p  I p R
• This is equation (33-5) in your text. To make this look like
Ohms law we define capacitive reactance. What is the
definition for capacitive reactance?
1
XC 
wC
• What are the units?
ohms

dQ d
I  I p sin wt 
 Qp sin wt  wQp cos wt
dt dt
Qp
Ip
VC , p 

 XCIp
C wC
AC Circuits
• Let’s check this out. Connect the 2.2 kW resistor on the E&M
board in series with the 0.047 mF capacitor to the output of the
function generator. Set the amplitude to 5 volts, the frequency to
200 Hz, and the function to sine wave. Connect the voltage probes
to the Pasco 750 Interface across the resistor. To get the phase
correct connect the black voltage lead to one side of the resistor and
connected the ground lead (black cable) from the ground side of the
signal generator. We will also measure the voltage across the
capacitor by connecting the leads from channel B across the
capacitor. Make sure the positive lead is connected to the positive
side of the capacitor. This is the side connected to the positive lead
from the signal generator. You need to click the plug icon and drag
it to channel B and choose Voltage Sensor. To connect the output to
the oscilloscope do not open a new window but go to the
oscilloscope window and click the second trace icon (Which should
be No Signal) on the right and select Channel B.
AC Circuits
red
red
channel A
channel B
C
ac emf
red
R
channel C
• Measure the peak voltage across the resistor and the
capacitor. Use the resistance to calculate the peak
current. Record the values in the table below. Change
the frequency and repeat the measurements.
frequency
voltage
voltage
current
capacitive
(resistor) (capacitor)
reactance
Hz
Vp (V)
Vp (V)
Ip (mA)
XC (W)
____20__ __________ __________ __________ __________
__2000__ __________ __________ __________ __________
AC Circuits
• Measure the peak voltage across the resistor and the
capacitor. Use the resistance to calculate the peak
current. Record the values in the table below. Change the
frequency and repeat the measurements.
f
1/f
V(cap)
V(res)
I
XC
50
0.02
5
0.16
7.27273E-05
68750
100
0.01
5
0.32
0.000145455
34375
300
0.003333
4.9
0.93
0.000422727
11591.4
600
0.001667
4.6
1.8
0.000818182
5622.222
1000
0.001
4.2
2.6
0.001181818
3553.846
1500
0.000667
3.6
3.4
0.001545455
2329.412
2000
0.0005
3
3.9
0.001772727
1692.308
3000
0.000333
2.4
4.3
0.001954545
1227.907
AC Circuits
• Use DataStudio or open Excel and plot the capacitive
reactance as a function of frequency. Does the curve
look like what you would expect from the definition of
capacitive reactance? Explain.
XC vrs f
1
1
XC 

wC 2fC
So it should
look like a 1/x
curve
Xc (ohms)
Yes!
80000
70000
60000
50000
40000
30000
20000
10000
0
XC
0
1000
2000
f (sec)
3000
AC Circuits
• Use the definition to figure out how to plot the data so
that you get a straight line. Replot the data and do a
linear fit. From the slope calculate the capacitance.
1 1

XC  
 2C  f
XC vrs 1/f
Xc (ohms)
So slope
equals 1/2C
80000
70000
60000
50000
40000
30000
20000
10000
0
1
5 1
 35 X 10
2C
F
1
C
F  0.045mF
6
0
0.005
2  3.5 X 10
XC
0.01
0.015
1/f (1/sec)
0.02
0.025
AC Circuits
• Next we will measure the phase shift between the voltage
across the capacitor and the current through the capacitor.
Set the frequency to 2000 Hz. Make sure both positive
voltage leads are connected to the side connected to the
positive side of the signal generator. The two traces
should look like Figure 33-4.
red
channel A
channel B
C
red
ac emf
red
R
channel C
AC Circuits
• Measure the phase difference between the current and
voltage. Do this by measuring the period which is 3600
and the time shift between the two signal. Then
q = 3600 t/T where t is the time shift and T is the period.
• Change the frequency and see if the phase difference
between the two signals changes.
T
t
AC Circuits
• Inductive Reactance
• When we discuss inductors in a circuit there are two
important points for an inductor. One is the phase
relationship between the current in and voltage across an
inductor. What is the phase relationship? This is the
bold statement at the bottom of page 836 in your text.
The voltage across an inductor leads the current by 900.
Voltage reaches a maximum at time T
Current reaches a maximum at a later time T+DT
AC Circuits
• The second point is the magnitude relationship between
current and voltage for a inductor. What is this
relationship?
VL, p  I p X L
• This is equation (33-7) in your text. To make this look
like Ohms law we define inductive reactance. What is
the definition for inductive reactance?
XL  wL  2fL
• What are the units?
ohms
dI
VL  VL , p sin wt   L 
dt
d
  L I p sin wt  wLI p cos wt
dt
AC Circuits
• On the axis below show what you expect for the inductive
reactance as a function of frequency.
X L(ohms)
f (1/sec)
AC Circuits
• Replace the capacitor with the inductor in the circuit.
Using the digital meter measure the voltage across the
inductor and resistor as we did for the capacitor.
red
red
channel A
L
channel A
red
R
channel B
Signal/Function Generator
frequency
voltage
voltage
current
capacitive
(resistor) (capacitor)
reactance
Hz
Vp (V)
Vp (V)
Ip (mA)
R (W)
____20__ __________ __________ __________ __________
__2000__ __________ __________ __________ __________
AC Circuits
• Measure the peak voltage across the resistor and the
inductor. Use the resistance to calculate the peak current.
Record the values in the table below. Change the
frequency and repeat the measurements.
f
1/f
V(res)
V(ind)
I
XL
50
0.02
4.8
0.5
0.0022
229.17
300
0.0033
4
1.26
0.0018
693
600
0.0017
3.9
1.55
0.0018
874.36
1000
0.001
3.7
2
0.0017
1189.2
2000
0.0005
3.2
3
0.0015
2062.5
3000
0.0003
2.7
3.7
0.0012
3014.8
4000
0.0003
2.2
4.1
0.001
4100
5000
0.0002
1.8
4.3
0.0008
5255.6
AC Circuits
• Plot the inductive reactance as a function of frequency.
Does the graph agree with what you know about inductive
reactance?
XL vrs f
Yes!
6000
XC  wL  2fL
XL (ohms)
So t he slope is 2L.
5000
XC  wL  2L f
4000
3000
XL
2000
1000
0
0
The inductance is
1.1W
 170mH
L
2
1000
2000
3000
f (1/sec)
4000
5000
6000
AC Circuits
• Measure the phase shift between the current and voltage
across the inductor. Your two traces should look like
Figure 33-6.
red
red
channel A
red
R
Signal/Function Generator
channel A
L
channel B
AC Circuits
Summary of AC circuit equations
This is everything you need to know from today
AC Circuits
Summary of AC circuit concepts
This is a little more than you want to know!
AC Circuits
• Phase shift and Phasors
Active and Reactive Power
When a circuit has resistive and reactive parts, the resultant power has 2 parts:
– The first is dissipated in the resistive element. This is the
active power,P
– The second is stored and returned by the reactive element. This is the
reactive power, Q , which has units of volt amperes reactive or var
While reactive power is not dissipated it does have an effect on the system
– for example, it increases the current that must be supplied and increases
losses with cables
Therefore
 Active Power
 Reactive Power

P = VI cos  watts
Q = VI sin  var
Apparent Power
 S2 = P2 + Q2
S = VI VA
Ohm’s Law in an AC Circuit
 rms values will be used when discussing
AC currents and voltages
– AC ammeters and voltmeters are designed to
read rms values
– Many of the equations will be in the same
form as in DC circuits
 Ohm’s Law for a resistor, R, in an AC
circuit
– ΔVrms = Irms R
• Also applies to the maximum values of v and i
Summary of Circuit Elements,
Impedance and Phase Angles
THANK YOU