Ultimo 17 supch10.pdf

CHAPTER 10 - SINUSOIDAL STEADY-STATE ANALYSIS

List of topics for this chapter :Nodal AnalysisMesh AnalysisSuperposition TheoremSource TransformationThevenin and Norton Equivalent CircuitsAC Op Amp Circuits

NODAL ANALYSIS

P r o b l e m 1 0 . 1P r o b l e m 1 0 . 1 Given the circuit in Figure 10.1 and )t1000sin(5)t(i = amps, find )t(v ousing nodal analysis.

Figure Figure 1010..11

Carefully DEFINE the problem.Carefully DEFINE the problem.Each component is labeled, indicating the value and polarity. The problem is clear.

PRESENT everything you know about the problem.PRESENT everything you know about the problem.The goal of the problem is to find )t(v o , which is clearly labeled in Figure 10.1, usingnodal analysis. Thus, we need to label the nodes and ground.

To find )t(v o without using derivatives and integrals, we must transform the circuit tothe frequency domain. This allows us to find the answer using algebra with complexnumbers. We can transform the circuit to the frequency domain after setting areference value. Let us use a reference of )t1000sin(A f+ .

In transforming to the frequency domain, remember that LjX L w= and)Cj(1X C w= . Hence, the inductor becomes 10j)1010)(10(jLj 3-3 ==w and the

capacitor becomes -j20)]1050)(10(j[1Cj1 6-3 ==w .

30 30 WW

10 10 WW

50 50 mmFF 10 mH10 mH++

vvoo(t)(t)

--

10 10 WW

i(t)i(t)

iioo(t)(t)

Let us draw the circuit after the transformation into the frequency domain and labeling thenodes and ground.

Establish a set of ALTERNATIVE solutions and determine the one that promises theEstablish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.greatest likelihood of success.The problem clearly states that the problem be solved using nodal analysis. Thus, thetechnique to solve the problem is set. There is no reason to look at an alternative at thispoint.

ATTEMPT a problem solution.ATTEMPT a problem solution.

Let us begin by writing the node equations.At node 1 : At node 2 :

010

VV30

0V5- 211 =

-+

-+ 0

10j100V

20j-0V

10VV 2212 =

+-

+-

+-

Simplifying,

150V3V3V 211 =-+ 020)j1(V

20Vj

10VV 2212 =

-++

-

150V3V4 21 =- 0)j1(VVj)VV)(2( 2212 =-++-0V3V2- 21 =+

Thus, the system of simultaneous equations is

=

0

150

V

V

32-

3-4

2

1

=

=

-=

50

75

300

450

61

0

150

42

33

)612(1

V

V

2

1

So,= 075V1 or )t1000sin(75)t(v1 = volts= 050V 2 or )t1000sin(50)t(v 2 = volts

Clearly,

oo I10jV =

5500 A A 30 30 WW

10 10 WW

j20 j20 WW j10 j10 WW++

VVoo

--

10 10 WWVV11 VV22

IIoo

and )45-2)(5.2()j1)(5.2(20

)j1)(50()j1)(10(

5010j10

VI 2o =-=

-=

+=

+= amps

Hence,

)45-2)(5.2)(9010(I10jV oo == == 4536.3545225 volts

Therefore,)45t1000sin(36.35)t(v o += volts

EVALUATE the solution and check for accuracy.EVALUATE the solution and check for accuracy.Solving the problem with an alternate method, such as mesh analysis in this case, wouldshow that the results of the problem solution are correct.

Let us draw the circuit defining the loop currents for mesh analysis.

Write the loop equations.Loop 1 : = 05I1 ampsLoop 2 : 0)II(20jI10)II(30 32212 =--+-Loop 3 : 0I10jI10)II(20j- 3323 =++-

Simplifying the equations for loops 2 and 3,Loop 2 : 0I20jI)20j40(I30- 321 =+-+Loop 3 : 0I)10j10(I20j 32 =-+

Solve the third loop equation for 3I in terms of 2I .

2223 I45-4142.1I)j1(Ij1010j20-

I =-=-

=

Now, substitute 1I and 3I into the second loop equation and simplify to find 2I .

0I)j1)(20j(I)20j40()0(-30)(5 22 =-+-+0I)20j20(I)20j40()0(-30)(5 22 =++-+

)05)(30(I60 2 =

=

= 05.260

)05)(30(I 2 amps

Now, find 3I .

5500 30 30 WW

10 10 WW

j20 j20 WW j10 j10 WW++

VVoo

--

10 10 WW

IIoo

II22II11 II33

=== 45-536.3)05.2)(45-4142.1(I)45-4142.1(I 23 amps

Clearly, == 45-536.3II 3o amps.

Evidently,=== 4536.35)45-536.3)(9010(I10jV oo volts

and )45t1000sin(36.35)t(v o += volts

This answer is the same as the answer obtained using nodal analysis. Our check foraccuracy was successful.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return ALTERNATIVE solutions and continue through the process again.This problem has been solved satisfactorily.

=)t(v o volts)45t1000sin(36.35 ++

P r o b l e m 1 0 . 2P r o b l e m 1 0 . 2 [10.5] Given the circuit in Figure 10.1 and )t1000cos(10)t(v i =volts, use nodal analysis to find )t(v o .


Let us start by building the ac circuit. The voltage source, )t1000cos(10)t(v i = volts,becomes 010 volts, with 1000=w . The inductive reactance becomes 10jLj =w . Thecapacitive reactance becomes -j20)]1050)(1000(j[1Cj1 6- ==w . Now, draw the ac

20 20 WW

+-

vvii (t)(t)

50 50 mmFF 10 mH10 mH

20 20 WW 30 30 WW++

vvoo(t)(t)

--

4i4ioo(t)(t)

iioo(t)(t)

VV22VV1120 20 WW

+-

j20 j20 WW j10 j10 WW

20 20 WW 30 30 WW++

VVoo

--

4I4Ioo101000 V V

IIoo

circuit.

At node 1 : At node 2 :

20j-VV

20V

20V10 2111 -+=

-10j30

V20V4

20j-VV 2121

++=

-

)VV(jVV10 2111 -+=- 21 V)8.0j6.0(V)j4-( +=+

21 Vj2jV)j2(10 +-+= 21 Vj4-8.0j6.0

V+

+=

Note that 20VI 1o = was substituted when writing the equation for node 2.

Substituting the equation for node 2 into the equation for node 1,

22 VjVj4-)j8.06.0)(j2(

10 -+

++=

or2.26j6.0

170V 2 -

=

Clearly,

=

-

+=

+= 26.70154.6

2.26j6.0170

j33

V10j30

30V 2o volts

With a reference of )t1000cos(A f+ ,=)t(v o volts)26.70t1000cos(154.6 ++

P r o b l e m 1 0 . 3P r o b l e m 1 0 . 3 Given the circuit in Figure 10.1 and )t1000cos(20)t(v = volts, find)t(i o using nodal analysis.


=)t(i o milliamps)18.44-t1000cos(5.632

8 8 WW

6 6 WW

10 mH10 mH

10 10 WW

20 20 WWv(t)v(t) +-

iioo(t)(t)

250 250 mmFF

MESH ANALYSIS

P r o b l e m 1 0 . 4P r o b l e m 1 0 . 4 Given the circuit in Figure 10.1 and )t1000cos(2)t(i = amps, find )t(i ousing mesh analysis.


First, we transform the circuit to the frequency domain using a reference of)t1000cos(A f+ and define the mesh currents as seen in the following circuit.

Remember that LjX L w= and Cj1

X C w= .

There is only one unknown loop current, oI .Writing the loop equation,

0I20jI10)2I(10 ooo =++-20I)20j20( o =+

=

=

+=

+= 45-7071.0

452

01j1

120j20

20I o amps

Therefore,=)t(i o milliamps)45t1000cos(1.707 --

i(t)i(t)

10 10 WW

10 10 WW 20 20 mHmH

iioo(t)(t)

IIoo

10 10 WW

10 10 WW j20 j20 WW2200 A A

P r o b l e m 1 0 . 5P r o b l e m 1 0 . 5 Given the circuit in Figure 10.1 and )t1000sin(5)t(i = amps, find )t(i ousing mesh analysis.


Transform the circuit to the frequency domain using a reference of )t1000sin(A f+ anddefine the mesh currents.

Clearly, 2o II = .

Use mesh analysis to find 1I and 2I .For loop 1 : 0)II(20jI10)5I(30 2111 =--+-For loop 2 : 0I10jI10)II(j20- 2212 =++-

Simplifying,150I20jI)20j40( 21 =+-0I)10j10(I20j 21 =-+

Simplifying further,5.7IjI)j2( 21 =+-

0I)j1(I2j 21 =-+

Find 1I in terms of 2I for the second loop equation.

2221 I)j1(21

I21

j21

I2j

j1-I +

=

+=

+

=

Substituting this equation into the first loop equation we get,

30 30 WW

10 10 WW

50 50 mmFF 10 10 mHmH

10 10 WW

i(t)i(t)

iioo(t)(t)

30 30 WW

10 10 WW

j20 j20 WW j10 j10 WW

10 10 WW

II11 II225500 A A

IIoo

5.7IjI2

j1)j2( 22 =+

+

-

15I2jI)j1()j2( 22 =++-15I)3j3( 2 =+

=

=

+=

+= 45-536.3

452

05j1

53j3

15I 2

Hence,= 45-536.3I o amps

Therefore, =)t(i o amps)45t1000sin(536.3 --

SUPERPOSITION THEOREM

The superposition theorem applies to ac circuits the same as it does for dc circuits.This theorem is important if the circuit has sources operating at different frequencies.Since the impedances depend on frequency, we must have a different frequency-domaincircuit for each source. The total response is obtained by adding the individual responses inthe time domain.

P r o b l e m 1 0 . 6P r o b l e m 1 0 . 6 Given the circuit in Figure 10.1, )t1000cos(2)t(i = amps and)t3/4000sin(10)t(v = volts, find )t(iC .


Because the two sources have different frequencies, we need to use superposition to solvethis problem. Thus, )t(i)t(i)t(i 2C1CC += .

Start with the current source and a reference of )t1000cos(A f+ .

+-

v(t)v(t)i(t)i(t) 50 50 mmFF

iiCC (t)(t)

20 20 WW

j20 j20 WW

IIC1C1

2200 A A 20 20 WW

Using current division,

==+=+

=-

=

-= 454142.1452j1

2)j1)(2(

j12

)2(20j20

20I 1C amps

Hence,)45t1000cos(4142.1)t(i 1C += amps

Next, use the voltage source with a reference of )t3/4000sin(A f+ .

Clearly,

=

=++

=-

=-

= 87.364.025

)87.365)(2(916

)3j4)(2(3j4

215j20

10I 2C amps

Hence,)87.36t3/4000sin(4.0)t(i 2C += amps

Recall that )t(i)t(i)t(i 2C1CC += .

Therefore,=)t(iC amps)]87.36t3/4000sin(4.0)t1000cos(4142.1[ ++++

P r o b l e m 1 0 . 7P r o b l e m 1 0 . 7 [10.33] Given the circuit in Figure 10.1 and )t3cos(12)t(v i = volts,solve for )t(v o using the principle of superposition.


+-

20 20 WW

101000 V V

IIC2C2

j15 j15 WW

1/12 F1/12 F+-

vv ii (t)(t)

6 6 WW 2 H2 H

4sin(2t) A4sin(2t) A +-

10 V10 V

++

vvoo(t)(t)

--

Let 321o VVVV ++= where 1V , 2V , and 3V are respectively the voltages produced bythe 10 volt dc source, the ac current source, and the ac voltage source acting independently.For 1V , consider the circuit shown in Figure (a).

The capacitor is an open circuit to dc while the inductor is a short circuit. Hence, 10V1 =volts and 10)t(v1 = volts.

For 2V , consider the circuit in Figure (b). 2=w , so the inductor becomes 4jLj =w .Likewise, the capacitor becomes 6j-)12/2(j-Cj1 ==w .

Applying nodal analysis,

2222 V

41

j61

j61

4jV

6j-V

6V

4

-+=++=

which leads to

=-

= 56.2645.21j2)12)(4(

V 2 volts

With a reference of )t2sin(A f+ ,)56.26t2sin(45.21)t(v 2 += volts

1/12 F1/12 F

6 6 WW 2 H2 H

+-

10 V10 V

++

VV11

--

((a)a)

((b)b)

j6 j6 WW

6 6 WW j4 j4 WW

++

VV22

--

4400 A A

For V3, consider the circuit in Figure (c). 3=w which leads to 6jLj =w for the inductorand -j4(3/12)j-Cj1 ==w for the capacitor.

At the non-reference node,

6jV

4j-V

6V12 333 +=

-

which leads to

=+

= 56.26-733.10j2)12)(2(

V3 volts

With a reference of )t3cos(A f+ ,)56.26t3cos(733.10)t(v 3 -= volts

Recall that )t(v)t(v)t(v)t(v 321o ++= .

Therefore,=)t(v o volts)]56.25t3cos(733.10)56.25t2sin(45.2110[ --++++++

P r o b l e m 1 0 . 8P r o b l e m 1 0 . 8 Given the circuit in Figure 10.1 and )t1000cos()t(v S = volts, find )t(v .


=)t(v volts)90t1000cos(2010 --++or =)t(v volts)t1000sin(2010 ++

((c)c)

j4 j4 WW+-

6 6 WW j6 j6 WW

++

VV33

--

121200 V V

+-

10 V10 V +-

vvSS (t)(t)

+ -v(t)v(t)

20 20 WW 50 50 mmFF

20 20 mHmH

SOURCE TRANSFORMATION

Source transformation in the frequency domain involves transforming a voltagesource in series with an impedance to a current source in parallel with an impedance or viceversa. We must keep the following relationship in mind when performing sourcetransformations.

SSS IZV =S

SS Z

VI =

P r o b l e m 1 0 . 9P r o b l e m 1 0 . 9 Given the circuit in Figure 10.1 and )t1000cos(20)t(v = volts, find)t(v o using source transformations.


Transform the circuit to the frequency domain using a reference of )t1000cos(A f+ .

Reduce the circuit using source transformations. Begin with the 200V source in serieswith 6W which becomes a 10/30A source in parallel with 6W.

8 8 WW

6 6 WW

10 mH10 mH

10 10 WW

20 20 WWv(t)v(t) +-

250 250 mmFF

++

vvoo(t)(t)

--

8 8 WW

6 6 WW

j10 j10 WW

10 10 WW

20 20 WW+-

j4 j4 WW

++

VVoo

--

202000 V V

8 8 WW j10 j10 WW

10 10 WW

20 20 WW

j4 j4 WW

++

VVoo

--

10/310/300 A A 6 6 WW

Now combine 6W || 8W = 24/7W. The 10/30A source in parallel with 24/7W becomes an80/70V source in series with 24/7W.

To perform the next source transformation, we need to find the parallel equivalent of theresistor and capacitor in series. We know that two series impedances and two parallelimpedances are

2S1SSeq ZZZ += and2P1PPeq Z

1Z1

Z1

+=

To find the parallel equivalence of two series impedances, let PeqSeq ZZ = . Then,

2P1P2S1S Z1

Z1

ZZ1

+=+

It can be shown that

j6.939-1

095.81

4j7241

+=-

Thus, the 80/70V source in series with 24/7 j4W becomes a 2.16949.4A source inparallel with 8.095 j6.939W.

Combining 8.095W || 20W = 5.763W. We now need to find the series equivalent of theresistor in parallel with the capacitor. It can be shown that

833.2j411.31

j6.939-1

763.51

-=+

Thus, the 2.16949.4A source in parallel with 5.763 j6.939W becomes a 9.6189.69Vsource in series with 3.411 j2.833W. Before redrawing the circuit, lets combine the seriesresistances of 3.411W + 10W = 13.411W.

j6.939 j6.939 WW j10 j10 WW

10 10 WW

20 20 WW++

VVoo

--

8.095 8.095 WW2.1692.16949.449.4AA

24/7 24/7 WW

j10 j10 WW

10 10 WW

20 20 WW+-

j4 j4 WW

80/780/700 V V++

VVoo

--

Now that we have found a less complex circuit via source transformations, it is simple tosolve for )t(v o .

=

=

+

=+-

= 43.18-6325.012.28206.15

69.9618.9167.7j411.13

69.9618.910j833.2j411.13

69.9618.9I

and=== 57.71325.6)43.18-6325.0)(9010(I10jVo

Recall that we used a reference of )t1000cos(A f+ . Therefore,=)t(v o volts)57.71t1000cos(325.6 ++

THEVENIN AND NORTON EQUIVALENT CIRCUITS

P r o b l e m 1 0 . 1 0P r o b l e m 1 0 . 1 0 Given the circuit in Figure 10.1 and )t1000cos(100)t(v = volts, find

ThV , NI , and eqZ looking into terminals a and b.


Because the circuit in Figure 10.1 has a capacitor, we can only determine ThV , NI , and eqZfor an ac circuit in the frequency domain. Hence, the circuit becomes,

j2.833 j2.833 WW13.411 13.411 WW

j10 j10 WW+-

9.6189.6189.699.69VV II++

VVoo

--

3i(t)3i(t)+-

5 5 WW

v(t)v(t)

i(t)i(t)

5 5 WW

1 mF1 mF

aa

bb

Begin by finding the open-circuit voltage, ocV . Note that there is no current flowing

through the capacitor. Thus, 1oc VV = .

Using nodal analysis,

0I35

0V5100V 11 =-

-+

-

where 5

V100I 1

-= .

So,

05

V100)3(

5

0V

5

100V 111 =

-

--

+-

0)V100)(3(V)100V( 111 =--+-400V5 1 =

80V1 = volts

Hence,= 080Voc volts

Now, find the short-circuit current, scI .

Use nodal analysis to find 1V and then

+-

5 5 WW

II

5 5 WW

j j WW

3I 3I10010000 V V

VV11

++

VVococ

--

+-

5 5 WW

II

5 5 WW

j j WW

3I 3I10010000 V V

VV11

IIscsc

j-V

I 1sc =

Writing the nodal equation,

0j-

0VI3

50V

5100V 111 =

-+-

-+

-

where 5

V100I 1

-= .

So,

0Vj5

V100)3(

5

0V

5

100V1

111 =+

-

--

+-

0V5j)V100)(3(V)100V( 1111 =+--+-400V)5j5( 1 =+

)j1)(40(2

)j1)(80(j1

805j5

400V1 -=

-=

+=

+=

Thus,

=+=-

-== 45240)j1)(40(

j)j1)(40(

j-V

I 1sc amps

Finally,

=

== 45-2

45240

080IV

Zsc

oceq ohms

Therefore,== ocTh VV 080 or =)t(v Th volts)t1000cos(80

== scN II 45240 or =)t(iN amps)45t1000cos(57.56 ++

=eqZ 45-2 or =eqZ ohms)j1( --

P r o b l e m 1 0 . 1 1P r o b l e m 1 0 . 1 1 [10.43] Find the Thevenin and Norton equivalent circuits for thecircuit shown in Figure 10.1.

5 5 WW j10 j10 WW

j20 j20 WW

2 2 WW

6060120120 V V +-


To find ThZ , consider the circuit shown in Figure (a).

=-=+-=+-= 69.33-63.2112j182)12j16(2)10j5(20jZeq ohms

To obtain ThV , consider the circuit in Figure (b).

+

=+-

= 120602j1

4j12060

20j10j520j

Voc

== 57.14633.107)12060)(57.267889.1(Voc volts

=

== 26.180961.469.33-63.2157.14633.107

ZV

Ieq

ocsc amps

where scI is the current flowing downward through a short across the terminals.

Recall that eqNTh ZZZ == , ocTh VV = , and scN II = .

Therefore,== NTh ZZ ohms)12j18( --

=ThV volts57.14633.107

ZZeqeq

(a)(a)

5 5 WW j10 j10 WW

j20 j20 WW

2 2 WW

(b)(b)

5 5 WW j10 j10 WW

j20 j20 WW

2 2 WW

6060120120 V V +-

++

VVococ

--

=NI amps26.180961.4

P r o b l e m 1 0 . 1 2P r o b l e m 1 0 . 1 2 Given the circuit in Figure 10.1 and )t1000cos(100)t(v = volts, find)t(i , the current through R, for WWWWW= 1000and,100,10,1,0R .


R I )t(i

W0 45-657.5 A A)45-1000tcos(657.5 W1 45-439.5 A A)45-1000tcos(439.5

W10 45-041.4 A A)45-1000tcos(041.4 W100 45-1314.1 A A)45-1000tcos(1314.1 W1000 45-138.0 A mA)45-1000tcos(138

AC OP AMP CIRCUITS

P r o b l e m 1 0 . 1 3P r o b l e m 1 0 . 1 3 Given the ac circuit in Figure 10.1, find outV as a function of inV .

+-

v(t)v(t)

10 10 WW 10 mH10 mH

50 50 mmFF

5 5 WW

RR

-+

VVaa

VVbb

j1000 j1000 WW

1 k1 kWW

+-

VVinin

++

VVoutout

--

10 10 WW


Using nodal analysis at node a,

010j-

VV10

VV3outa

3ina =

-+

-

where 0VV ba == .

So,

010j-V-

10V-

3out

3in =+

inout VVj- =

inin

out Vjj-V

V ==

Therefore,=outV 90Vin

The output is equal to the input except for a phase shift of 90.

P r o b l e m 1 0 . 1 4P r o b l e m 1 0 . 1 4 Given the circuit in Figure 10.1 and )tsin(10)t(v in w= volts, find)t(v out for 1000,100,10,1=w rad/sec.


-+

VVaa

VVbb

10 10 mmFF

10 k10 kWW

+-

vv iinn (t)(t)

++

vvoutout(t)(t)

--

10 10 WW

10 H10 H

Use the following ac circuit, with a reference of )tsin(A f+w , to find outV in terms of w.

Using nodal analysis at node a,

010j

VV10j-

VV10

10V outa5

outa4

a =w

-+

w-

+-

where 0VV ba == .

So,

010jV-

10j-V-

1010- out

5out

4 =w+

w+

0V10j

Vj100- out4

out =w+w-

100V10j

j- out4

=

w+w

-w

w=

-ww

=w+w

= 9010

10010

100j10jj-

100V 42424out

Now, substitute the values for w into the equation for outV .

At 1=w rad/sec, =@ 90-01.09010-100

V 4out

=)t(v out mV)90tsin(10 --

At 10=w rad/sec, ==-

@ 90-10101.0909900-10

901010

10V

3

42

3

out

10 10 WW

-+

VVaa

VVbb

j10j1055 //ww WW

101044 WW

+-

++

VVoutout

--

j10j10ww WW

101000

=)t(v out mV)90t10sin(01.101 --

At 100=w rad/sec, =-

@ 900

1090

101010

V4

44

4

out

=)t(v out This corresponds to the case where the LC combination forms a parallel resonant circuitand the output goes to infinity.

At 1000=w rad/sec, =

=-

@ 9010101.090109.9

1090

101010

V 55

46

5

out

=)t(v out mV)90t1000sin(01.101 ++

In conclusion, the output, )t(v out , has a 90 phase shift for all values of w less than 100 andhas a 90 phase shift for values of w greater than 100.

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PrefaceChapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of AnalysisChapter 4 Circuit TheoremsChapter 5 Operational AmplifierChapter 6 Capacitors and InductorsChapter 7 First-Order CircuitsChapter 8 Second-Order CircuitsChapter 9 Sinusoids and PhasorsChapter 10 - Sinusoidal Steady-State AnalysisNodal AnalysisMesh AnalysisSuperposition TheoremSource TransformationThevenin and Norton Equivalent CircuitsAC OP Amp Circuits

Chapter 11 AC Power AnalysisChapter 12 Three-Phase CircuitsChapter 13 Magnetically Coupled CircuitsChapter 14 Frequency ResponseChapter 15 Laplace TransformChapter 16 Fourier SeriesChapter 17 Fourier TransformChapter 18 Two-Port Networks

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Ultimo 17 supch10.pdf