1

Basic Circuit Laws

Units

2

1Nm 1Ws=

1F 1s/=

1H 1s = ⋅

3

SI Prefixes

4

Independent Sources

Ideal voltage source Ideal current source

Ideal assumption: no resistive element in sourceRealistic assumption:

RR

The internal resistor of an ideal voltage source is zero !The internal resistor of an ideal current source is infinity !

Current source has an internal parallel resistor

Voltage source has an internal series resistor

5

Ohm’s LawRA

=

{ }1

( ) ( ) Ohm's Law( ) [ ] Resistance( )1 ( ) [S] Conductance

( )

[ , , ]

v t R i tv tRi t

i tGR v t

mho

-

= ⋅

=

= =

[Siemens]

Instantaneous Power2

2 2( )( ) ( ) ( ) ( ) ( )v tp t v t i t R i t G v tR

= ⋅ = ⋅ = = ⋅

Note: - p(t) is a parabolic (non-linear) function that is always positive.- p(t) is no indicator for the direction of power flow.

6

Examples:

0Box absorbes power (resistor)p v i v R i= ⋅ > = ⋅

0Box provides power (source)p v i v R i=- ⋅ > =- ⋅

0Box provides power (source)p v i v R i=- ⋅ > =- ⋅

( ) 0 ( ) ,Box absorbes power (resistor)p v i v R i v R i=- ⋅ - > - = ⋅ - = ⋅

7

Kirchhoff’s Current Law (KCL)

2i

1R 2Rs

1 2

1 2

1 2

00

s

s

s

i i ii i i

i i i

- - =

- + + =

= +

- The algebraic sum of currents entering a node is zero.- The algebraic sum of currents leaving a node is zero.- The algebraic sum of currents entering a node equals the sum of

currents leaving the node.

The algebraic sum of all currents at any node in a circuit equals zero.0

In general: 0N

nn

i=

=å

Example: For the above circuit, find the equivalent resistance.

8

Example - continued1 2

1 21 2

1 21 2

1 2 1 2

1 2

Note:

1 1 1

s

s ss

seq

s eq

seq

s

v v vv vi i iR R

i R Ror R R Rv R R R R R

i G G Gv

= =

= + = +

⋅= + = = =

+

= + =

2i

1R 2Rs

In general:

1

1

1 1n

jeq j

n

eq jj

R R

G G

=

=

=

=

å

å

99

Kirchhoff’s Voltage Law (KVL)

1 2 1 2

0

0 or 0

In general: 0

s sN

nn

v v v v v v

v=

- + + = - - =

=åThe algebraic sum of all voltages around any closed loop in a circuit equals zero.

Example: Find the equivalent resistance.

1 2 1 2 1 2

1 2

0 ( )s s s

seq

v v v v R i R i v R R iv R R Ri

- - = - ⋅ - ⋅ = = + ⋅

= + =

In general:

1

n

eq jj

R R=

=å

10

Voltage Divider1

11 2 1 2

1 22 1 2

21 2

s

s s

s

Rv vR R R Rv v v v

R R Rv vR R

üïï= ïï+ +ï + = =ýï +ï= ïï+ ïþ

Current Divider

1 1 2 1 21

1 2 2 1 1 2

2 2 1 2 12

1 2 1 2 1 2

11 1 1

11 1 1

s s s stot

s s s stot

G R R R Ri i i i iG R R R R R RG R R R Ri i i i iG R R R R R R

üïï= = = = ïï+ + + ïýïï= = = = ïï+ + + ïþ

1 21 2

1 2s s

R Ri i i iR R

++ = =

+

11

SuperpositionThe principle of superposition states that whenever a linear system isdriven by more than one independent source, the total response can befound by summing the individual responses to each independentsource.When applying this principle, short-circuit a voltage source, and opena current source.

Example: Find the voltage across the 3 resistor.

3v+-

Step 1: Deactivate current source (voltage divider) Example: Find the voltage across the 3 resistor.

3vsv +-

33 (2 4 )120V

6 3 (2 4 )2120V 30V

6 2

vsv

+=

+ +

= =+

Example: Find the voltage across the 3 resistor.

12

Step 2: Deactivate voltage source (current divider)

3csv+- [ ]

[ ] [ ]

''3

''3

12 (3 6 )

112A4 2 (3 6 )

4 2 (3 6) 4 2 2 22 (3 6) 2 2 4

6A

i

i

+=

+

+ += = =

+ +

=''

'' ''22 3''

3

12 2 23 4A

1 1 2 1 3 33 6

ii i

i

=- =- =- =- =-++

''3 23 3 ( 4A) 12Vcsv i = ⋅ = ⋅ - =-

Step 3: Superposition

3 3 3 30V 12V=18Vvs csv v v= + = -

13

Mesh-Current Analysis

R = Resistance matrix

1414

Example: Find the matrix system of this circuit using mesh analysis.

1

2

3

4

5

170 40 0 80 0 2440 80 30 10 0 0

V0 30 50 0 20 1280 10 0 90 0 100 0 20 0 80 10

iiiii

é ùé ù é ù- - ê úê ú ê úê úê ú ê ú- - - ê úê ú ê úê úê ú ê ú⋅ =- - -ê úê ú ê úê úê ú ê ú- - ê úê ú ê úê úê ú ê ú- -ê ú ê ú ê úë û ë ûë û

15

Node-Voltage Analysis

G = Conductance matrix

16

Example: Find the matrix system of this circuit using node analysis.

1 2 3

4

1

2

3

4

1 2 1 1 1 1 0 0 6A1 1 1 1 1 1.2 1 4 1 1.2 1 4 0

S0 1 12 1 1.2 1 8 1 12 0 00 1 4 0 1 4 1 3 1 6 0

vvvv

é ùé ù é ù+ - ê úê ú ê úê úê ú ê ú- + + - - ê úê ú ê ú⋅ =ê úê ú ê ú- + + ê úê ú ê úê úê ú ê ú- + +ê ú ê úê úë û ë ûë û

17

Nodal Versus Mesh Analysis

Given a network to be analyzed, how do we know which method is better or more efficient? The choice of the better method is dictated by two factors.

Networks that contain many series-connected elements, voltage sources, or supermeshes are more suitable for mesh analysis.

Networks with parallel-connected elements, current sources, or supernodes are more suitable for nodal analysis.

Also, a circuit with fewer nodes than meshes is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is better analyzed using mesh analysis.

The key is to select the method that results in the smaller number of equations.

If node voltages are required, it may be expedient to apply nodal analysis.

If branch or mesh currents are required, it may be better to use mesh analysis.

Mesh analysis is the only method to use in analyzing transistor circuits, but mesh analysis cannot easily be used to solve op amp circuits.

For non-planar networks, nodal analysis is the only option, because mesh analysis only applies to planar networks.

18

Thevenin Equivalent Circuit

Thevenin’s theorem states that a linear two-terminal circuit can bereplaced by an equivalent circuit consisting of a voltage source VThand a resistor RTh, where VTh is the open-circuit voltage at theterminals and RTh is the input or equivalent resistance at the terminalswhen the independent sources are turned off.

19

Norton Equivalent Circuit

Norton’s theorem states that a linear two-terminal circuit can bereplaced by an equivalent circuit consisting of a current source IN anda resistor RN, where IN is the short-circuit current at the terminals andRN is the input or equivalent resistance at the terminals when theindependent sources are turned off.

20

Example: Find the Thevenin voltage and Thevenin resistance.

Solution: Node equation

1 1

1

25V3A 0

5 2032V=ab Th

v v

v v V

-+ - =

= =

Node equation for

2 2 2

2

25V3A+ 0

5 20 416V16V 4A4sc

v v v

v

i

-+ - =

=

= =

2v

32V 84A

ThTh Th

sc

VR R

i= = =

21

Source Transformations

ss

viR

=s sv R i= ⋅

Maximum Power Transfer 22

2 2

max since 24

æ ö÷ç ÷= =ç ÷ç ÷ç +è ø

= = =

ThL L L

Th L

ThTh

Th L

Vp I R R

R R

V Vp V V

R R

Maximum power is transferred to the load when the load resistanceequals the Thevenin (Norton) resistance.

22

Note: Later in the course we will deal with alternating currents andvoltages, leading to complex impedances. In this case, we have

ThZ

*=L ThZ Z

2 2

max1Maximum power: 4

= =Th

Th L

V Vp

Z Z

2

max,av 2=

L

Vp

Z

The average power (for (AC) is related to the rms value:

23

Example: Find the value of RL for maximum power transfer in the following circuit. Find the maximum power..

Solution:

Th6 122 3 (6 12 ) 5 9

6 12R ⋅

= + + = + =+

24

Example - cont’d

Maximum Power: 2 2Th

maxTh

(22V) =13.44 W4 4 9VpR

Mesh equation: 1 2 2 112V 18 12 0 , 2A 2 3Ai i i i - + ⋅ - ⋅ = =- =

KVL around the outer loop:

1 212V 6 3 2 (0) 0 22VTh Thi i V V - + ⋅ + ⋅ + ⋅ + = =

25

Example: Determine the value of RL that will draw the maximum power from the rest of the circuit. Calculate the maximum power.

Note: In order to find the Thevenin resistance of circuits with dependent sources, we need to excite the circuit at its terminals.

26

Example - cont’d

We need to find RTh and VTh. To find RTh, we consider the circuit in Fig. (a).

Fig. (a) Fig. (b)

0 0 0

0

0 0 00

0

Applying KVL at the top node:1V 3

4 1 2But . Hence1V 4 1 V

4 1 2 191V 1 1 19 9A A

4 4 38381V 4.2229

x

x

Th

v v v v

v vv v v v

vi

R i

0 0

0

0 0 0 0

0

2 2

max

To find V , consider the circuit in Fig. (b):9V+2 +1 3 0

But 2 . Hence9V=3 +6 9 1A

9V 2 7V4.222

(49V) 2.901W4 4 4.222

Th

x

x

Th

L Th

Th

L

i i vv i

i i i iV iR R

VPR

27

Example: Is the 6V source absorbingpower and, if so, how much?

1.) 40V 5 8A= 2.) 5 20 4 , 4 8A 32V = ⋅ =

3.) (6 4 10) 20 ,4 8A 32V 20 1.6A

+ + =⋅ = =

4.) 20 30 12 , 12 1.6A 19.2V = ⋅ =

i

6V 6V19.2V 6V 0.825A , 6V 0.825A 4.95W

16i p v i

-

= = = ⋅ = ⋅ =

Solution:

Note: Source transformation in this way for independent sources only !

28

Delta-to-Wye (-to-T) Equivalent Circuits

A circuit viewed as a circuit.

A Y circuit viewed as a T circuit.

29

Transformation <> Y

30

Capacitor

A typical capacitor

A capacitor with applied voltage v.

[ ]sF

qCv

31

KCL =>

The equivalent capacitance of n parallel-connected capacitors is the sum of the individual capacitances.

3232

KVL =>

The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual

capacitances.

33

Inductor

A typical inductor

Li

[ ]H s

34

KVL =>

The equivalent inductance of series-connected inductors is the sum of the individual inductances.

35KCL =>

The equivalent inductance of parallel inductors is the reciprocal of the sum of the reciprocals of the individual inductances.

36

Inductor and Capacitor Comparison

Inductor Capacitor

Symbol

Units Henries [H]=[s] Farads [F]=[s/]

Describing equation

Other equation

Initial condition i(to) v(to)

Behavior with const. source

If i(t) = I, v(t) = 0 short circuit

If v(t) = V, i(t) = 0 open circuit

Continuity requirement

i(t) is continuous so v(t) is finite

v(t) is continuous so i(t) is finite

dttdiLtv )()(

t

t oo

tidvL

ti )()(1)(

dttdvCti )()(

t

t oo

tvdiC

tv )()(1)(

37

Inductor and Capacitor Comparison cont’dInductor Capacitor

Power

Energy

Initialenergy

Trapped energy

Series-connected

Parallel-connected

dttditLititvtp )()()()()(

221 )()( tLitw

221 )()( oo tLitw

221 )()( Liw

)()(221

ooeq

eq

titiLLLL

)()()()(

1111

321

321

ooooeq

eq

titititiLLLL

dttdvtCvtitvtp )()()()()(

221 )()( tCvtw

221 )()( oo tCvtw

221 )()( Cvw

)()(221

ooeq

eq

tvtvCCCC

)()()()(

1111

321

321

ooooeq

eq

tvtvtvtvCCCC