Circuits

Exam 1

Summer 2020

1. /25

2. /30

3. /30

4. /15

Total /100

Name __________________

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Problem 1) KCL/KVL (25 pts)

1.1: 20 pts Using KCL and K VL and ohm's law only, find the voltages for all resistors. There aretwo spaces for you to fill in to give the value of R2 and R3 before you start this problem. You MUSTchoose an integer 1-9, for every space you see. For example, for R2 if I choose 1 for the first spaceand 2 for the second space, the value of R2 is 1.2K ohms.

___.____K ohms ___.____K ohms

2

My solution:

AB

C

ILeft

IRight

Iouter is not drawn in to avoid clutter. Also all reference marks are top down + - respectively

or left right + - respectively.

R1 2200 R2 1400 R3 2300 R4 3500

V1 8 I1 0.003 V2 2

You will find that you cannot use KVL loops and one KCL loop to solve this problem. The matrix will not converge. So you need two KCL loops. With recognition of V2 and R3 in series (able to switch around), two KCL loops can be found considering IR3 for nodes A and C. It looks as if we are breaking a rule of not using node with a voltage source at first. This concept will be revisited in the conceptual questions.

KCL at node C

iR1 iR2 iR3 0=

VR1

R1

VR2

R2

VR3

R3 0=

3

For KVL around the left loop

8 VR1 VR2 VR4 0=

2( ) VR1 VR2 VR4 8=

For KVL around right loop.

VR2 2V VR3 0=

3( ) VR2 VR3 2=

KCL at node A:

i2 i3 i4 0.003=

VR2

R2

VR3

R3

VR4

R4 0.003=

4( )VR2

R2

VR3

R3

VR4

R4 0.003=

C1

0

8

2

0.003

M1

1

R1

1

0

0

1

R2

1

1

1R2

1

R3

0

1

1

R3

0

1

0

1

R4

M11

C1

5.941

3.107

1.107

1.048

4

Conceptual 1.2: 5 pts

___.____K ohms ___.____K ohms

ORIGINAL CIRUCIT

Circle the following circuits that willgive the same answers for voltagesacross each resistor. For everycircuit you circled explain why youchose it. Even if you choosecorrectly, not explaining your answerwill be marked incorrect.

Explaination (if chosen:) Explaination (if chosen):

Explaination (if chosen): Explaination (if chosen):

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A (things in series can be switched)CD (things in parallel can be switched)

Problem 2) Nose/Mesh Analysis (30 points)

FOR THIS PROBLEM YOU MUST PICK DIFFERENT VALUESTHAN WHAT YOU CHOSE IN PROBLEM 1! There are twospaces for you to fill in to give the value of R2 and R3 beforeyou start this problem. You MUST choose an integer 1-9, forevery space you see.

___.____K ohms ___.____K ohms

2.1: 15 pts Use NODAL analysis to find the voltage across R4.

Students don't have to calculate the number of equations needed for nodal but I did.

Node KCL equations 5 2 1 2

6

AB

C

ILeft

IRight

D

Node KCL equations 5 2 1 2

R1p2 2200 R2p2 3500 R3p2 4900 R4p2 3500

V1p2 8 I1p2 0.003 V2p2 2

From node A

I1

VA

R4p2

VA VC

R2p2

VA VD

R3p2 0=

VA1

R4p2

1

R2p2

1

R3p2

VC1

R2p2

VD1

R3p2

0.003=

From supernode VCD

VR4 (V)

7

VC VB

R1p2

VC VA

R2p2

VD VA

R3p2 0=

VA1

R2p2

1

R3p2

VC1

R1p2

1

R2p2

VD

R3p2 8

1

R1p2

=

Supernode definition

VC VD 2=

M2

1

R4p2

1

R2p2

1

R3p2

1

R2p2

1

R3p2

0

1

R2p2

1

R1p2

1

R2p2

1

1

R3p2

1

R3p2

1

VA 2.513

C2

I1p2

8

R1p2

2

M21

C2

2.513

2.98

0.98

VC 2.98

VD 0.98

VB 8V=VR4 VA 2.513

VD VA 3.493

2.2 15 pts With the same values for R2 and R3 from 2.1, use MESH analysis to find the voltageacross R3.

8

AB

C

i1

i3

D

i2

KVL equations 3 1 2

For supermesh loop i1 and iw

8 i2 R1p2 i2 R2p2 i3 R2p2 i1 R4p2 0=

i1 R4p2 i2 R1p2 R2p2 i3 R2p2 8=

For mesh loop 2

i3 R2p2 i2 R2p2 V2 i3 R3p2 0=

i2 R2p2 i3 R2p2 R3p2 2=

supermesh definition

i2 i1 3mA=

9

M3

R4p2

0

1

R1p2 R2p2

R2p2

1

R2p2

R2p2 R3p2

0

i1 7.18 104

C3

8

2

0.003

M3

1C3

7.18 104

2.282 103

7.127 104

i3 7.127 10

4

VR3 i3 R3p2 3.492

VR4p2 i1 R4p2 2.513

Problem 3) Thevenin Dependent Circuits (30 pts)

FOR THIS PROBLEM YOU MUST PICK DIFFERENT VALUESTHAN WHAT YOU CHOSE IN PROBLEM 1 and PROBLEM 2!There are two spaces for you to fill in to give the value of R2and R3 before you start this problem. You MUST choose an

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integer 1-9, for every space you see. There is a dependent source in this problem!

___.____K ohms ___.____K ohms

R3 is the Load‐

+Vx

3Vx

3.1: 10 pts R3 is the load. Find VTH using the Open Circuit method. Hint: You will find mesh

analysis to be easier. You can choose nodal if you'd like, however.

R1p3 2200 R2p3 4700 R3p3 1500 R4p3 3500

V1p3 8 I1p3 0.003 V2p3 2

Take of the load. Can use any analysis. I used mesh.

supermesh KVL

8 i1 R4p3 i2 R1p3 i2 R2p3 0=

1( ) i1 R4p3 i2 R1p3 R2p3 8=

supermesh definition

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i1 i2 3Vx=

i1 i2 3Vx 0=2( )

dependent source definition

i1 R4p3 Vx=

3( )i1 R4p3 Vx 0=

C4

8

0

0

M4

R4p3

1

R4p3

R1p3 R2p3

1

0

0

3

1

i1p3 1.104 107

M41

C4

1.104 107

1.159 103

3.865 104

i2p3 1.159 103

Vx1 3.865 104

VTh (V)

VTH i1p3 R4p3 i2p3 R2p3 2 3.447

3.2: 10 pts Find INorton uisng short circuit current (Isc).

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R3 is shorted so add i3

supermesh KVL

8 i1 R4p3 i2 R1p3 i2 R2p3 i3 R2p3 0=

1( ) i1 R4p3 i2 R1p3 R2p3 i3 R2p3 8=

supermesh definition

i1 i2 3Vx=

i1 i2 3Vx 0=2( )

dependent source definition

IN (ohms)i1 R4p3 Vx=

3( )i1 R4p3 Vx 0=

i3 loop

i3 R2p3 i2 R2p3 2 0=

4( ) i2 R2p3 i3 R2p3 2=

C5

8

0

0

2

M5

R4p3

1

R4p3

0

R1p3 R2p3

1

0

R2p3

R2p3

0

0

R2p3

0

3

1

0

i3p3 2.302 103M5

1C5

2.598 107

2.728 103

2.302 103

9.093 104

iSC i3p3 2.302 103

for future calcuation checkVTH

iSC1.497 10

3

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3.3. 8 pts Find Rth using the Test Method.

Similar calcuation as above but with a 1V source where R3 is. Short all independent sources.

supermesh KVL

i1 R4p3 i2 R1p3 i2 R2p3 i3 R2p3 0=

1( ) i1 R4p3 i2 R1p3 R2p3 i3 R2p3 0=

supermesh definition

i1 i2 3Vx=

i1 i2 3Vx 0=2( )

dependent source definition

i1 R4p3 Vx=

3( )i1 R4p3 Vx 0=

i3 loop

i3 R2p3 i2 R2p3 1 0=

4( ) i2 R2p3 i3 R2p3 1=

M6

R4p3

1

R4p3

0

R1p3 R2p3

1

0

R2p3

R2p3

0

0

R2p3

0

3

1

0

C6

0

0

0

1

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iTest 6.674104A

M61

C6

4.33 108

4.546 104

6.674 104

1.516 104

RTh (Ω)

RTH1V

iTest1.498 10

3 Ω

3.4: 2 pts Verify your answers from 3.1-3:3. (Verify they are correct).

VTH

INRTH=

VTH

iSC1.497 10

3 close rounding errors

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4) Cascading Multi-stage Op Amp Circuits and Design Concepts (15 pts)

4.1: 10 pts You have 3 inputs, V1, V2, and V3. Design and draw a cascade of op amps thatresult in the equation -3V1 + 9V2 - 6V3. You are limited to 4 op amps. Make sure to add in yourresistor values.

Graders there is more than one way to dothis problem. As long as it makes sense...

4.2: 5 pts You have a resistive sensor that must set off an alarm if the sensor resistance isbetween 100 ohms and 1k ohms. The alarm is triggered between 1V and 3.5V whichcorresponds with these resistance values. You have a access to +/- 5V sources for this circuit.

Using only high level building blocks, show how you would create this circuit. Every buildingblock must include an input and output as shown below. If additional power is needed into thebuidling block, indicate this as well.

1. Choose any resistive sensor2. Use building blocks learned in unit 1 that help determine the resistance

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3. Use buillding blocks learned in unit 1 that set thresholds4. Use building blocks learned in unit 1 that makes decisions5. Choose the output you'd use and any associated components that may be needed to make itwork

_______Input

(i.e. vibration)

Output

(i.e. Resistance change)

Building block circuit from

Unit 1

sensors can be:temperature sensorlight sensor (ldr)pressure sensorstrain gaugeanything that changes resistance

to determine resistance:voltage dividerbridge circuit

to set threshold:reference voltage dividers

to make decisions:comparator 1 comparator 2 window comparatorschmidtt trigger

to alarm:LED (should include current limiting resistor but ok notincluded)Buzzerother...

For each of these, they should have input and output that make sense...linput light output resistance changethen output voltage then output set threshold voltage then output saturated voltage then current limit if using a current limiting resistor then output light on from LED for example...

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