review psc kcl and kvl

8/4/2019 Review Psc Kcl and Kvl

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1

Electric circuitan interconnection of circuit elements.There must be a closed loop for the current to flow.

+

+

There is a closed loop, enabling

current to flow into resistorNo closed loop


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2

Circuit elementeach component of a circuit.

* A two terminal electrical device, it can be completely

characterized by its voltage and current relationship

-active elements generate energy (voltage and current sources)

-passive elements cannot generate energy (resistors, capacitors,inductors)

This course deals with:1. Power for an element

2. Voltage across an element

3. Current through an element


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3

Voltage is the energy needed to move a unit charge between 2 points in space orin a circuit. It is expressed in volts (V).

Voltage must be an oriented variable. The (+) and (-) signs indicate voltage polarity:

+ vab -

a b- vba +

* Above, point a is vab potential higherthan point b.

* voltage rise from b to a is vab* Since voltage vba is defined with the opposite orientation, it has theopposite value of voltage vab.

vba = - vab

dq

dwvab

dw is energy in joules required to move

a negative charge dq from a to b


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4

Voltage

Lowercase letter represents general case

v may or may not be time varying

Uppercase letter if quantity is not time varying

V is not time varying (DC Voltage)

Lowercase letter followed by a (t) is time varying

v(t) is time varying (AC Voltage)


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5

Electric current is the flow of electric charges through space per unit time.It is measured in ampheres (A).

current, positive charges

electrons

Current is oriented with a directed arrow

Current through an object is the same going in, as it is coming out of the objectChanging the orientation of current changes its sign (see below)

dt

dqi

i i

-i-i


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6

Current

Lowercase letter represents general case

i may or may not be time varying

Uppercase letter if quantity is not time varying

I is not time varying (DC Current)

Lowercase letter followed by a (t) is time varying

i(t) is time varying (AC Current)


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Passive Sign Convention

Passive Sign Convention is satisfied when current enters through the + terminal

of an element and leaves through theterminal.

Circuit Element

E1

+ i1v1

PSC NOTPSC

E2

+ v2 i2

E3+

i3

v3

PSC

i1

Pabs = i vWhen PSC is satisfied

When PSC is NOT satisfied Pabs = - (i v)

Power absorbed by an element

Power absorbed by device = - Power delivered by

device / supplied by device

i2


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PSC Examples

Given v1 = - 3 volts, i1 = 3 amps

What is PABS_E1?

Notice: v1 and i1 are not PSC

But ix and v1 are PSC

ix = - i1 = - (3 amps)

Therefore, PABS_E1= ix v1= (-3 amps)(-3 volts)= 9 watts

Similarly, PABS_E1= -(i1 v1 )= -(3 amps)(-3 volts)= 9 watts

E2

+

vx

ix Given vx = 10 Volts, ix = - 4 ampsWhat is PABS_E2?

vx and ixfollow PSC

PABS_E2= ix vx= (- 4 amps)(10 volts)= - 40 watts


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But what is a circuit element?

A Circuit Element is mathematical model of a 2 terminal electrical device, it can

be completely characterized by its voltage and current relationshipResistors:

R

i

+

-

01

vR

i

y = mx + b

R>0

i

v

1/R

Current voltage

characteristic

equation

Ohms law variationsv = i R

R= , i 0

i

v

Unit of resistance is ohm ()

Note: The above relations assume PSC as we

have drawn, if i and v do not follow PSC for aresistor then v = - i R


10/4010

Ohms Law and PSC

PSC is satisfied:given i = 2mA, R = 5 K

R1

i+

-

v = i R = (2mA)(5K) = 10 V

PABS_R1 = i v = (2mA)(10V) = 20 mW

Note: if i and v do not follow PSC for a resistor then v = - i R

+

-

iPSC is not satisfied:given i = 4A, R = 2v = -(i R) = - (4A)(2) = - 8 V

PABS_R2 = -( i v) = -(4A)(-8V) = 32W

R2


11/4011

Open Circuit

i

v

As R increases the line flattens out

As R the function approaches any voltage, but

i=0

Open Circuitno path for current to flow

i=0

+

-

vR =


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Short Circuit

i

v

As R0 the line gets vertical

At R=0 we have a Short Circuit

v = iR = 0

i can be anything, but v=0

i+

-

R = 0 v = 0


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Examples Short Circuit/Open Circuit

13

Short circuit between terminals a and b

Open circuit between terminals c and d


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14

Independent Voltage SourcesOrientation indicated by polarity inside bubble

Current assigned follows PSC

i

v+

i

v(t)+

i

VB

+

-

V indicates a constantsource like a battery

v(t) indicates a time varying source(typically sinusoidal)

i

v

Current-voltage

characteristics

3

i

v=V=3 volts+

Current is unknown until you

find out what is connected to

the source!


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15

Independent Current Sources

Orientation indicated by arrow inside.

Guarantees a motion of charge but musthave path to flow and must be connected to

an object.v+

-

i, I, i(t)

i

v

3

v

+

-

I=3A

Example: constant current source

Current is fixed at 3 A.

Voltage can be anything!

Note: Ohms law does not apply to current or voltage sources.

Also, current and voltage sources can apply OR absorb power to a circuit.


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16

Circuit Terminology

A node is the point of connection of one or more circuit

elements If a short circuit (a wire) connects 2 nodes, they constitute a single

node

A loop is any closed path (starting & ending at the same

location)How many nodes and loops in the circuits below?

+

+

+


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17

Circuit Terminology

+

+

+

2 nodes

1 loop2 nodes

3 loops

3 nodes

1 loop


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Series/Parallel Resistors ?

19

36 V

10

20

30


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Series/Parallel Resistors ?

20

36 V

10

20

30

No resistors in series

20 and 30 ohm resistances are in parallel


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22

More on Power

As in our example, power absorbed by a resistor must be positive!

VR = I RTherefore PABS_R = IV = I I R = I

2R 0

Conservation of Energy or Power Rule

The sum of all the power absorbed by all N elements tied together in a circuit must equal

zero!

From previous example, PABS_IS + PABS- R = -5/2W + 5/2W = 0 Watts

Power Delivered

Power delivered by an element is equal to the opposite of the power absorbed by that same

element

PDEL_VS = - PABS_VS

0_1

iABS

N

i

P


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23

Example 2

R=15 Find the voltage, current and power absorbed for

each element. Note: it is often easier to work with

resistors first!

Is = 3A

R=15Is = 3A

Make the arbitrary assignments shown in red font

IR = - 3 Amps

To apply Ohms law, use current following PSC (IS)

VR = ISR = (3A)(15) = 45 V

Note: PABS_R = ISVR = (IS)2R = (VR)

2/R

Because I = (1/R)V and IV= (1/R)VV= V2/R

PABS_R = ISVR =(3 A)(45 V) = 135 W

VS = - VR = - (45 V) = - 45 V

PABS_IS = ISVS = ( 3 A)(-45 V) = - 135 Watts

+

-

VS

+

-

VR

IR

0_1

iABS

N

i

P PABS_IS + PABS_R = -135 W + 135 W = 0 Watts


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24

Example 3: Using Conservation of Energy to find unknown current

(assume you dont know KCL)

0__ VSABSISABSABS PPP

10 VoltsIs = 5A

+

-

IVS=?

PABS_IS = -(5A10V) = - 50 Watts

PABS_VS = - PABS_IS = -( - 50 Watts) = 50 Watts

PABS_VS = IVSVS

IVS = = 50W/10V = 5 Amps

Find IVS

S

VSABS

V

P _

+

-

10V

-

10V

+


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28

Resistors in Series

In the circuit to the right, we wish to replace

resistors R1, R2, and R3 with one

Equivalent resistance!

For resistors in series:

RTOTAL=R1+R2+R3

N

N

i

iTOTAL RRRRR ...21

1

+

R1 R2

VS R3

Node A

Node B

+

VS RTOTAL

A

B


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29

Resistors in Parallel

In the circuit to the right, we wish to replace

resistors R1, R2, and R3 with one

Equivalent resistance!

For resistors in parallel:

Notation: R1//R2//R3

Note: if we had just two resistors R1 and R2 in parallel,

we could then apply the product-sum rule, that is:

NN

i i

TOTAL

RRR

R

R

/1.../1/1

1

1

1

21

1

+

R1 R3

VS R2

321 /1/1/1

1

RRRRTOTAL

Node A

Node B

+

RTOTAL

A

B

VS

21

2*1

RR

RRRTOTAL


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30

Introduction to Current Divider

Is R2R1

I1 I2

R3

I3

How do we findI1,I2, andI3?

VS

+


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31

KCL: IS= I1 +I2 +I3 ,1

1

R

VI

S ,

2

2R

VI

S

3

3

R

VI

S

IsR2 VS

+

R1

I1 I2

R3

I3

321321

111

RRRV

R

V

R

V

R

VI S

SSSS

321

111

1

RRR

IV SS

Solve for Vs

But we know:

+

VS


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32

IsR2 VS

+

R1

I1 I2

R3

I3+

VS

321

111

1

RRR

IV SS

Current Division

Equation for the above

circuit topology

321

1

11

1

111

1

1

RRR

RIVRR

VI SSS


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33

Current Divider

This leads to a general current divider equation for N parallel

resistors

The above equation assumes that source current and

current through the circuit element flow in oppositedirections (one into the node, one out of the node)

N

i i

j

Sj

R

RII

1

1

1

C Di id Diff O i i f


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34

Current DividerDifferent Orientations of

element currents

BA

ASB

RR

RII

Is RBRA

IA IB

VS

+

BA

BS

BA

ASA

RR

RI

RR

RII

11

1

IC

ID

BA

BSC

RR

RII

BA

ASD

RR

RII

But, if we define the currents in the opposite direction we get:


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35

(KVL) Kirchhoffs Voltage Law

*Sum of voltages around a closed path (loop) in a circuit

is zero.A closed pathmove from point to point, only the first

and last points repeat.

Voltage Polarity Conventionwhen using KVL

A voltage encountered from + to - is positive

A voltage encountered from - to + is negative


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36

KVL

+

-

+

+

+

+

+

-

-

--

-

+

-

V1V2V3

V5V4

V6V7

A

F

C

B

D

E

Closed Path:

A B C D E A+V1 +V6 +V5 V4 V3 = 0

Another Closed Path:

E A F C D E-V3 -V2 +V7 +V5 V4 = 0


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37

KVL

+

-

+

+

-

-

-V1

V3 V2

A

C

BD

Closed Path:

B C D A B+(V2 )(V3)(-V4)+(V1) = 0

Use parenthesis, especially for

negative voltage variablesV4

andV1.

-V4+

-


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KVL Example 1

38

Clockwise Starting at A:

+ V1V2 + (- V3) = 0+(2V) -(V2) +(-(8V)) = 0

V2 = - 6V

V3 = 8 Volts

V1 = 2 Volts

find V2


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KVL Example 2

39

36 V

20

10

+

+ -

-

12 V

V2

Clockwise Starting at the bottom node:

- 36V +12V - V2 = 0

V2 = - 24 V

find V2


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KVL Example 3

find VZ

Clockwise Starting at positive terminal of 10 V:

+10V -5V -12V + VZ = 0

VZ = 12V + 5V -10V = 7V

review psc kcl and kvl

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