Drawn in impedance formssawyer/CircuitsSummer2020_all/... · 2020. 4. 21. · 0.003H 1 2 L2 0.3H 1...

Electric CircuitsECSE 2010

HW08 solution Spring 2020

Problem 1) Ideal Transformers and Power

V1

FREQ = 60HzVAMPL = 100V

VOFF =

R1

25

R2100

L1

0.398H

1 2

L20.266H

1

2

C1

0.53E-4F TX1

N 2.5=

The above circuit has a 100 V, 60 Hz source.

VA1 100V:= R1 25Ω:= C1 0.53 104−

F:= R2 100Ω:=

f 60Hz:= L2 0.266H:= L1 0.398H:=

ω 2 π f:=N1 2.5:=

ω 376.9911

s=

Drawn in impedance form1−

C1 ω 50.049− Ω=L2 ω 100.28 Ω= L1 ω 150.042 Ω=

ZL2 100jΩ:= ZL1 150jΩ:=ZC1 50j− Ω:=

V1

FREQ = 60HzVAMPL = 100V

VOFF =

R1

25

R2100

L1

150j

1 2

L2j100

1

2

C1

-50j TX1

Convert to thevenin using a voltagedivider (for voltage across L2...) Combine secodary impedances Note: This software

program uses i as thedefault character forimaginary numbers.Replace with j.VTh VA1

ZL2

R1 ZL2+:=

ZLoad ZL1 R2+:=

ZLoad 100 150i+( ) Ω=VTh 94.118 23.529i+( ) V=

Find thevenin impedance by shorting source then combining like resistors

1



ZTh

R1 ZL2

R1 ZL2+ZC1+:=

ZTh 23.529 44.118i−( ) Ω=

V1

FREQVAMPL = 94+23j

VOFF =

TX1

23.5-44j 100+150j

a. Determine the equivalent circuit when referring the secondary to the primary (and draw the circuit).

ZLEQsp

ZLoad

N12

:= Zssp ZTh:= Vssp 96.9 14deg<:=

Zssp 23.529 44.118i−( ) Ω= 942

23.52+ 96.893=

ZLEQsp 16 24i+( ) Ω=atan

23.5

94

14.036 deg=

V1

FREQ

VAMPL = 96<14deg

VOFF = 23.5-44j16 + 24j

2



b. Determine the equivalent circuit when referring the primary to the secondary (and draw the circuit).

Zsps N12

ZTh:= ZLEQps ZLoad:=

ZLEQps 100 150i+( ) Ω=Zsps 147.059 275.735i−( ) Ω=

Vsps N1 VTh:=

Vsps 235.294 58.824i+( ) V=

2352

58.82+ 242.245=

atan58.5

235.5

13.95 deg=

242.245 13.95deg<

V1

FREQ

VAMPL = 242.2 < 13.95 deg

VOFF = 147 - 275.7j100 + 150j

c. Determine the total complex power dissipated through the impedance in the primary.

Use the expression

Sprimary IRMS2

Zprimary cosθ primary j IRMS2

Zprimary sinθ primary +=

ZTh 23.529 44.118i−( ) Ω=Remember Z primary is

Need to find IRMS so we need to find the current in the primary circuit in part a.

Find ZEQ

ZEQ Zssp ZLEQsp+:=

3



ZEQ 39.529 20.118i−( ) Ω=

Convert to phasor/polar form

39.52

20.1−( )2+ 44.32=

ZEQphasor 44.3 27− deg<:=atan

20.1−39.5

26.97− deg=

Vssp 96.9 14deg<:=96.9

44.32.187=

Isp96.9 14deg<

44.3 27− deg<=

14 27+ 41=

Isp 2.187 41<=

Find phasor/polar form for impedance in the primary

Zssp 23.529 44.118i−( ) Ω=

23.52

44.1−( )2+ 49.971=

atan44.1−

23.5

61.948− deg=

Zsspphasor 50 62− deg<:=

IRMS2.187

2:=

Sprimary P jQ+= where

P2.187

2

250 cos 62− deg( ):=

P 56.137=

Q2.187

2

250 sin 62− deg( )

:=

4



Q 105.578−=

Sprimary 56.1 j 105.6−=

d. Determine the total complex power dissipated through the impedance in the secondary.

S IRMS2

Zsecondary cosθ secondary j IRMS2

Zsecondary sinθ secondary +=

From the circuit in part b.

ZEQ2 Zsps ZLEQps+:=

ZEQ2 247.059 125.735i−( ) Ω=

Convert to polar form

2472

125.7−( )2+ 277.145=

ZEQ2 277 26.8− deg<=atan

125−247

26.843− deg=

Vsps 242.245 13.95deg<= 242.245

2770.875=

Ips242.245 13.95deg<

277 26.8− deg<=

13.95 26.8+ 40.75=

Ips 0.875 40.75<=

ZLoad 100 150i+( ) Ω=The impedance in the secondary is

1002

1502+ 180.278=

atan150

100

56.31 deg= Zloadphasor 180.3 56.3deg<=

IRMS0.875

2:=

5



Ssecondary P jQ+= where

P0.875

2

2180.3 cos 56.3deg( ):=

P 38.296=

Q0.875

2

2180.3 sin 56.3deg( )

:=

Q 57.422=

Ssecondary 38.3 j 57.4+=

e. Verify that the power dissipated is equal to the power generated by the source.

Ssource

VRMS2

ZEQcosθ ZEQ j

VRMS2

ZEQsinθ ZEQ

+=

Vsource 96.9 14deg<=

VRMS96.9

2:=

Ssource P j Q( )+=

remember from above in part c.

ZEQphasor 44.3 27− deg<:=Psource

96.9

2

2

44.3cos 27− deg( ):=

Psource 94.427=

Qsource

96.9

2

2

44.3sin 27− deg( ):=

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Qsource 48.113−=

SSource 94.6 j 48.1−=

The complex power of the Secondary and Primary circuits should add up to the Source power.

Ssecondary 38.3 j 57.4+= Sprimary 56.1 j 105.6−=

SSecondary Sprimary+ 38.3 j 57.4+ 56.1+ j 105.6−=

56.1 38.3+ 94.4= 105.6− 57.4+ 48.2−=

SSecondary Sprimary+ 94.4 j 48.2−=

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Problem 2) Real Transformers

V1

FREQ =VAMPL = 5V

VOFF =

R1

25

1:4

R210k

a. Determine the voltage across R2 in phasor form

Refer the primary to secondary

R12 25Ω:= R22 10kΩ:=Vs 5V:=

N2 4:=

Vsps2 N2 Vs:=

Vsps2 20 V=

Zsps2 N22

R12:=

Zsps2 400Ω=

V1

FREQ =VAMPL = 5V

VOFF =

R1

25R210k

L10.003H

1

2

L20.3H

1

2

M k L1 L2=

b. If the real transformer is constructed of two inductors (as show) and the coupling coefficient is k=0.8,determine the Tee model of the circuit. Draw your circuit.

k 0.8:=

L12 0.003H:=

L22 0.3H:=

M k L12 L22:=

8



M 0.024 H=

In the Tee model, the inductor would be replaced with

L12 M− 0.021− H=

L22 M− 0.276 H=

M 0.024 H=

V1

FREQ =VAMPL = 5V

VOFF =

R1

25

R210k

L1-M

-0.021H

1 2L2-M

0.276H

1 2

M0.276H

1

2

9



Problem 3) Impedance matching

V1

FREQ = 1kVAMPL = 20V

VOFF =

R1

50

R250

1:5

C13.18E-7

10:1

R3

7.5

L10.00398

1

2

Z1

a. For the transformer circuit abive, design Z1 such that the power dissipated across the resistor R3 is maximized.

L13 0.00398H:= f3 1kHz:=N2a 5:= R13 50Ω:= C13 3.18 10

7−F:= ω3 2 π f3:=

N2b1

10:= R23 50Ω:= 1−

C13 ω3500.487− Ω= ω3 L13 25.007 Ω=

V1 20V:= R33 7.5Ω:=ZL13 25jΩ:=

ZC13 500j− Ω:=

Taking the 'left' circuit, refer the primary to the secondary ('left' to 'center')Taking the 'right' circuit, refer the secondary to the primary ('right' to 'center')

First find the thevenin equivalent...

VTh3

R23

R13 R23+V1:=

ZTh3 25Ω:=VTh3 10 V=

Refer to secondary

Vsps3 N2a VTh3:=

Vsps3 50 V=

Zsps3 N2a2

ZTh3:=

Zsps3 625Ω=

10



Refer to primary

ZLEQ R33 ZL13+:=

ZLEQsp3

ZLEQ

N2b2

:=

ZLEQsp3 750 2.5i 103+( ) Ω=

V1

FREQ = 1kVAMPL = 50

VOFF =

R1

625

C1-j500

R3

750

L1j2500

1

2

Z1

The impedance to the left of the dashed line needs to be the comple conjugate of the impedance to the right ofthe dashed line to be maximized

R13 Z1+( ) C13

R13 Z1+ C13+750 j2500−=

625 Z1+( ) 500j−

625 Z1+ 500j−+750 2500j−=

42625

73−

44500

73i+

44500

73609.589=42625−

73583.904−=

Z1 584− j610+=

Note: Negative resistance is only possible with dependent sources. With only passive comonents, the best thatcould be done is Z1 = j610 (inductor).

11



Problem 4) Mutual Inductance

L10.01

1 2

L20.01

1 2

R150

L30.04

1 2

R2100

R31k

R4

25V1FREQ = 159VAMPL = 2VOFF = 0

V2

FREQ = 318VAMPL = 2VOFF = 0

L40.1

1 2M2 = 0.01

M1 = 0.005

Find the current and voltage through R1 in the above couple circuits. There are two locations with inductive coupling andboth sets have additive coupling. Additionally, there are two voltage sources with different excitation frequencies.Suggestion, use superposition in your analysis.

12



Use the T-model for mutual inductance. Note, one of the equivalent inductances is 0. This is equivalent to a shortcircuit.

M42 0.01:=M41 0.005:=

L43 0.04:= L42 0.01:=

L41 0.01:= L44 0.1:=

L42 M42− 0=L43 M41− 0.035=

L41 M41− 5 103−= L44 M42− 0.09=

R1100

R21k

R3

25 R450

L10.005

1

2

L20.035

1

2

L30

1

2

L40.09

1

2

L5

0.01

12

L6

0.005

12

V1FREQ = 159VAMPL = 2VOFF = 0

V2


USE SUPERPOSITION!

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Cosidering the 159 Hz source

100

1k

25 50

j51

2

j351

2

j901

2

j10

12

j5

12


Using current divider circuit analysis to find IR1

Top100 35j+( ) 5j

100 35j+ 5j+0.216 4.914i+=

Bottom 1000 90j+( ) 10j1000 90j+ 10j+

0.099 9.99i+=

Add up impedances

25 0.216+ 4.914j+ 50+ 5j+ 0.099+ 9.99j+ 75.315 19.904i+=

IR12

75.315 19.904i+( ):=

IR1 0.02482 0.00656i−= [A]

0.024822

6.56− 103−( )2

+ 0.02567=

atan6.56− 10

3−0.02482

14.805− deg=

VR1 IR1 50:=

VR1 1.241 0.328i−= [V] at 159 Hz

1.2412

0.328−( )2+ 1.284=

VR1 in polar form for159 Hz source

atan6.56− 10

3−0.025

14.703− deg=

14



Considering the 318 Hz SourceYou can use current divider or maybe meshanalysis...I show both below

100

1k

25 50

j101

2

j701

2

j1801

2

j20

12

j10

12

V3


Using current divider:

Top100 70j+( ) 10j

100 70j+ 10j+0.61 9.512i+=

0.61 9.512i+ 10i+ 50+ 25+ 75.61 19.512i+=

Add up impedances in series (furthest away from source)

75.61 19.512i+( ) 20j75.61 19.512i+ 20j+

4.155 17.828i+=

2

4.155 17.828i+ 180i+ 1000+1.917 10

3− 3.777i 104−−=

To find IR1 in second loop away from source

20i

75.61 19.512i+ 20i+( )1.917 10

3− 3.777i 104−−( ) 2.866 10

4− 3.573i 104−+= [A] at 318 Hz

15



OR: Using mesh analysis, YAAAY!

i1 100 i1 j70+ i1 j10+ i2j10− 0=

i1 100 j80+( ) i2 j10− 0=

i2 j10 i1 j10− i2 j10+ i2 50+ i2 j20+ i3 j20− i2 25+ 0=

i1− j10 i2 j40 75+( )+ i3 j20− 0=

i3 1000 i3 j20+ i2 j20− i3 j180+ 2−=

i2− j20 i3 j200 1000+( )+ 2−=

M1

100 80i+( )

10i−

0

10i−

75 40i+( )

20i−

0

20i−

1000 200i+

:=

IR1CM

0

0

2

:=M1

1−CM

7.806− 106− 3.491i 10

5−+

2.867 104− 3.574i 10

4−+

1.917 103− 3.777i 10

4−−

= IR2

IR3

IR1bmag 2.867 104−( )2

3.574 104−( )2

+ 4.582 104−=:=

IR1bϕ atan3.574 10

4−

2.867 104−

51.264 deg=:=

[V]VR1b 2.867 10

4− 3.574i 104−+( ) 50 0.014335 0.01787i+=:=

VR1bmag 0.014335( )2

0.01787( )2+ 0.023=:=

VR1bϕ atan0.01787

0.014335

51.264 deg=:=

IR1Total 0.0257 cos 1000t 14deg−( ) 4.582 104− cos 2000t 51.24deg+( )+= [A]

VR1Total 1.28 cos 1000t 14deg−( ) 0.023 cos 2000t 51.24+( )deg[ ]+= [V]

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Drawn in impedance formssawyer/CircuitsSummer2020_all/... · 2020. 4. 21. · 0.003H 1 2 L2 0.3H 1...

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Transcript of Drawn in impedance formssawyer/CircuitsSummer2020_all/... · 2020. 4. 21. · 0.003H 1 2 L2 0.3H 1...