Vadodara Institute of Engineeringkotanbi-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 CircuitsAC Signals and rms Values

• There are a number of ways to describe the current or voltage for a time varying signal: peak value, peak-to-peak, 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?

• How is Vrms related to Vm?

We will derive this in a minute.

When you measure an ac current or voltage with a DMM you are measuring rms values.

mV

rmsV2

aveV 0

AC Circuits• Connect the cables from the Signal/Function Generator and the

voltage input cables to the 2.2 k 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(t)

• 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.

Hint: What is T?

• Finally, dividing by T and takethe square root you should get

Vrms Vp

2

dttVT

p )(sin 2

0

2

V2(t) = Vp2sin2(t)

Vp2 T

2

sin(2T )

4

Vp

2 T

2

T

p

T

p

ttVdttV

0

2

0

22

4)2sin(

2)(sin

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

AC CircuitsResistance

• 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 CircuitsCapacitive 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.

Voltage reaches a minimum at a later time T+T

Current reaches a minimum at time T

AC Circuits• The second point is the magnitude relationship between

current and voltage for a capacitor. What is this relationship?

• 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?

• What are the units?

XC 1

C

VC ,p Ip XC or VC ,rms IrmsXC

ohms

Like

VR ,p IpR

pC

pp

pC

ppp

IXC

I

C

QV

tQtQdtd

dtdQ

tII

,

cossinsin

AC Circuits• Let’s check this out. Connect the 2.2 k resistor on the E&M

board in series with the 0.047 F 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

• 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 () ____20__ __________ __________ __________ __________ __2000__ __________ __________ __________ __________

C

R

ac emfchannel A

channel B

channel C

red

red

red

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.

Yes!

fCCX C 2

11

So it should look like a 1/x curve

XC vrs f

01000020000300004000050000600007000080000

0 1000 2000 3000

f (sec)

Xc

(oh

ms)

XC

XC vrs 1/f

01000020000300004000050000600007000080000

0 0.005 0.01 0.015 0.02 0.025

1/f (1/sec)

Xc

(oh

ms)

XC

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.

fCX C

12

1

So slope equals 1/2C

FFX

C

FX

C

045.0105.32

1

11035

21

6

5

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.

C

R

ac emfchannel A

channel B

channel C

red

red

red

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

= 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+T

• The second point is the magnitude relationship between current and voltage for a inductor. What is this relationship?

• 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?

• What are the units?

VL,p Ip XL

XL L 2fL

AC Circuits

ohms tLItIdtd

L

dtdI

LtVV

pp

pLL

cossin

sin,

• On the axis below show what you expect for the inductive reactance as a function of frequency.

AC Circuits

X (ohms)L

f (1/sec)

• 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.

frequency voltage voltage current capacitive

(resistor) (capacitor) reactance

Hz Vp (V) Vp (V) Ip (mA) R ()

____20__ __________ __________ __________ __________

__2000__ __________ __________ __________ __________

AC Circuits

L

R

Signal/Function Generator

channel A

channel A

channel B

red

red

red

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

0

1000

2000

3000

4000

5000

6000

0 1000 2000 3000 4000 5000 6000

f (1/sec)

XL

(oh

ms)

XL

Yes!

XC L 2fL

So t he slope is 2L.

XC L 2L f

mH17021.1

L

The inductance is

AC Circuits• Measure the phase shift between the current and

voltage across the inductor. Your two traces should look like Figure 33-6.

L

R

Signal/Function Generator

channel A

channel A

channel B

red

red

red

AC CircuitsSummary of AC circuit equations

This is everything you need to know from today

AC CircuitsSummary of AC circuit concepts

This is a little more than you want to know!

AC Circuits• Phase shift and Phasors

Active and Reactive PowerWhen 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 P = VI cos watts

Reactive Power Q = VI sin var

Apparent Power S = VI VA

S2 = P2 + Q2

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

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