# survival skills for circuit analysis

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Survival Skills forCircuit Analysis

What you need to know from ECE 109Phyllis R. Nelson

prnelson@csupomona.edu

Professor, Department of Electrical and Computer Engineering

California State Polytechnic University, Pomona

P. R. Nelson Fall 2010 WhatToKnow - p. 1/46

mailto:prnelson@csupomona.edu

Basic CircuitConcepts

All circuits can be analyzed by manymethods.

P. R. Nelson Fall 2010 WhatToKnow - p. 2/46

Basic CircuitConcepts

All circuits can be analyzed by manymethods.

Some methods have specific advantages forcalculating a particular result in a specificcircuit.

P. R. Nelson Fall 2010 WhatToKnow - p. 2/46

Basic CircuitConcepts

All circuits can be analyzed by manymethods.

Some methods have specific advantages forcalculating a particular result in a specificcircuit.

Some methods yield useful intuition aboutcircuit behavior.

P. R. Nelson Fall 2010 WhatToKnow - p. 2/46

Basic CircuitConcepts

All circuits can be analyzed by manymethods.

Some methods have specific advantages forcalculating a particular result in a specificcircuit.

Some methods yield useful intuition aboutcircuit behavior.

No single analysis method will be thebest for all possible circuits!

P. R. Nelson Fall 2010 WhatToKnow - p. 2/46

Kirchhoffs Laws

These two methods are the most general tools ofcircuit analysis.

Sign errors in writing Kirchhoffs laws are themost common error I observe in student work inhigher-level classes.

Sign errors are not trivial errors!

P. R. Nelson Fall 2010 WhatToKnow - p. 3/46

Kirchhoffs current lawis equivalent to stating that electrical chargecannot be stored, created, or destroyed locally ina conductor.

The algebraic sum of all currentsentering a node is zero.

Alternate statements:

The algebraic sum of the currents leaving anode is zero.

The total current entering a node is equal tothe total current leaving the node.

P. R. Nelson Fall 2010 WhatToKnow - p. 4/46

Kirchhoffs voltage law

is derived by considering the forces on anelectrical charge that travels around a closedloop in a circuit and has the same starting andending velocity. The total force on the chargemust be zero, so the total change in electricalpotential energy (voltage) around the loop mustbe zero.

The algebraic sum of the voltagedrops around any closed loop in a

circuit is zero.

P. R. Nelson Fall 2010 WhatToKnow - p. 5/46

Solving K*L equations

Kirchhoffs laws applied directly result in

i

Ii = 0 (KCL)

i

Vi = 0 (KVL)

P. R. Nelson Fall 2010 WhatToKnow - p. 6/46

Solving K*L . . .

Solving KCL means finding the node voltages.

Solving KVL means finding the branch currents.

need relationships between currentand voltage for each circuit element.

These relationships describe the operation of thecircuit elements.

P. R. Nelson Fall 2010 WhatToKnow - p. 7/46

Ohms law+ V

I V = IR

Ohms law is an example of a current-voltage(I-V ) relation.

To use K*L, you must know the I-V relationfor every circuit element.

I-V relations include definition of therelationship between the direction of thecurrent and the sign of the voltage.

P. R. Nelson Fall 2010 WhatToKnow - p. 8/46

Series resistors,voltage divider

R1+ V1

R2+ V2

Ri+ Vi

Rn+ Vn

VT+I

VT = I Req

Prove that these equations are true:

Req =n

j=1

Rj Vi =Ri

nj=1 Rj

VT

P. R. Nelson Fall 2010 WhatToKnow - p. 9/46

VT =n

j=1

Vj =n

j=1

I Rj = In

j=1

Rj = I Req

Req =n

j=1

Rj

Vi = I Ri =

(

VTReq

)

Ri =

(

Rin

j=1 Rj

)

VT

P. R. Nelson Fall 2010 WhatToKnow - p. 10/46

Parallel resistors,current divider

V

+

I

R1 I1 R2 I2 V = I Req

Prove that these equations are true:

R1eq =2

i=1

R1i Req =R1R2

R1 + R2Ix =

(

RyRx + Ry

)

I

Which can be extended to more resistors?P. R. Nelson Fall 2010 WhatToKnow - p. 11/46

I = I1 + I2 =V

R1+

V

R2= V

(

2

i=1

R1i

)

=V

Req

R1eq =2

i=1

R1i

Req =1

R1+

1

R2=

R1R2R1 + R2

P. R. Nelson Fall 2010 WhatToKnow - p. 12/46

Let x = 1 and y = 2. Then

I1 =V

R1=

I ReqR1

=

(

R1R2R1+R2

)

R1I =

(

R2R1 + R2

)

I

If x = 2 and y = 1 then

I2 =

(

R1R1 + R2

)

I

P. R. Nelson Fall 2010 WhatToKnow - p. 13/46

For n resistors in parallel,

I =n

j=1

Ij =n

j=1

V

Rj= V

(

n

j=1

R1j

)

=V

Req

R1eq =n

j=1

R1j

This is the only formula for parallel resistors thatextends easily to more than two resistors.

P. R. Nelson Fall 2010 WhatToKnow - p. 14/46

Illegal circuits

The following configurations are not allowed:

a short-circuited voltage source

an open-circuited current source

Why?

P. R. Nelson Fall 2010 WhatToKnow - p. 15/46

KCL Example

+Vx

R1

R2

R3

R4 Iy

Find VR2 and VR4.

P. R. Nelson Fall 2010 WhatToKnow - p. 16/46

KCL Example

+Vx

R1

R2

R3

R4 Iy

Find VR2 and VR4.

Hint: Since KCL is applied at nodes, how manyKCL equations will be required?

P. R. Nelson Fall 2010 WhatToKnow - p. 16/46

KCL Example - 1

+Vx

R1

R2

R3

R4 Iy

Find VR2 and VR4.

P. R. Nelson Fall 2010 WhatToKnow - p. 17/46

KCL Example - 1

+Vx

R1

R2

R3

R4 Iy

Find VR2 and VR4.

1. Since KCL is applied at nodes, count thenodes.

P. R. Nelson Fall 2010 WhatToKnow - p. 17/46

KCL Example - 1

+Vx

R1

R2

R3

R4 Iy

Find VR2 and VR4.

1. Since KCL is applied at nodes, count thenodes. (4)

P. R. Nelson Fall 2010 WhatToKnow - p. 17/46

KCL Example - 2, 3

+Vx

R1

R2

R3

R4 Iy

Find VR2 and VR4.

P. R. Nelson Fall 2010 WhatToKnow - p. 18/46

KCL Example - 2, 3

+Vx

R1

R2

R3

R4 Iy

Find VR2 and VR4.

2. Choose one node as ground. Choose forconvenience!

3. Label node voltages.

P. R. Nelson Fall 2010 WhatToKnow - p. 18/46

+Vx

R1

R2

R3

R4 Iy

V2 V4

P. R. Nelson Fall 2010 WhatToKnow - p. 19/46

+Vx

R1

R2

R3

R4 Iy

V2 V4

Both VR2 and VR4 are connected to ground.

The voltage source is connected to ground one less KCL equation.

P. R. Nelson Fall 2010 WhatToKnow - p. 19/46

KCL Example - 4

+Vx

R1

R2

R3

R4 Iy

V2 V4

2 unknown node voltages 2 KCL equations.

4. Define currents for all elements connected tonodes 2 and 4.

P. R. Nelson Fall 2010 WhatToKnow - p. 20/46

KCL Example - Eqns

+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

KCL at 2:

P. R. Nelson Fall 2010 WhatToKnow - p. 21/46

KCL Example - Eqns

+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

KCL at 2:I1 I2 I3 = 0

P. R. Nelson Fall 2010 WhatToKnow - p. 21/46

KCL Example - Eqns

+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

KCL at 2:I1 I2 I3 = 0

KCL at 4:

P. R. Nelson Fall 2010 WhatToKnow - p. 21/46

KCL Example - Eqns

+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

KCL at 2:I1 I2 I3 = 0

KCL at 4:I3 I4 Iy = 0

P. R. Nelson Fall 2010 WhatToKnow - p. 21/46

KCL Example - I-Vs

+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

Write unknown Is in terms of node Vs:

P. R. Nelson Fall 2010 WhatToKnow - p. 22/46

KCL Example - I-Vs

+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

Write unknown Is in terms of node Vs:

I1 =Vx V2

R1I2 =

V2R2

I3 =V2 V4

R3I4 =

V4R4

P. R. Nelson Fall 2010 WhatToKnow - p. 22/46

KCL Example -Substitution

Vx V2R1

V2R2

V2 V4

R3= 0

V2 V4R3

V4R4

Iy = 0

Check! 2 equations in 2 unknowns

This pair of equations can be solved, but itsmessy unless resistor values are given.

P. R. Nelson Fall 2010 WhatToKnow - p. 23/46

KVL Example

+VA

R1

R2R3

I3

IB R4

Find I3.

P. R. Nelson Fall 2010 WhatToKnow - p. 24/46

KVL Example

+VA

R1

R2R3

I3

IB R4

Find I3.

Hint: Since KVL is applied around closed loops,how many KVL equations will be required?

P. R. Nelson Fall 2010 WhatToKnow - p. 24/46

KVL Example - 1

+VA

R1

R2R3

I3

IB R4

Find I3.

P. R. Nelson Fall 2010 WhatToKnow - p. 25/46

KVL Example - 1

+VA

R1

R2R3

I3

IB R4

Find I3.

1. Count the number of windowpanes in thecircuit.

P. R. Nelson Fall 2010 WhatToKnow - p. 25/46

KVL Example - 1

+VA

R1

R2R3

I3

IB R4

Find I3