E E 1205 Circuit Analysis

Lecture 2 - Circuit Elements and Essential Laws

Five Fundamental Elements

• Ideal Voltage Sources– Independent– Dependent

• Ideal Current Sources– Independent– Dependent

• Resistors• Inductors (to be introduced later)• Capacitors (to be introduced later)

Independent Voltage Source

• Voltage may be constant or time-dependent

• Delivers nominal terminal voltage under all conditions

Vg

Positive Terminal

Negative Terminal

Independent Current Source

• Current may be constant or time-dependent

• Delivers nominal terminal current under all conditions

Ig

Negative Node

Positive Node

Voltage-Controlled Dependent Voltage Source

• Terminal voltage is a function of the voltage drop of a different branch

• Delivers nominal terminal voltage under all conditions

v

Positive Terminal

Negative Terminal

+v-

Current-Controlled Dependent Voltage Source

• Terminal voltage is a function of the current flow in a different branch

• Delivers nominal terminal voltage under all conditions

i

Positive Terminal

Negative Terminal

i

Voltage-Controlled Dependent Current Source

• Current is a function of the voltage drop of a different branch

• Delivers nominal terminal current under all conditions

v

Positive Node

Negative Node

+v-

Current-Controlled Dependent Current Source

• Source current is a function of the current flow in a different branch

• Delivers nominal terminal current under all conditions

i

Positive Node

Negative Node

i

Electrical Resistance (Ohm’s Law)

• Electrical resistance is the ratio of voltage drop across a resistor to current flow through the resistor.

• Polarities are governed by the passive sign convention.

R

+ v -

i

vR

i

Power Consumed by Resistors

• Resistors consume power.

• v and i are both positive or both negative.

R

+ v -

i

p v i

v R i 2p i R

vi

R

2vp

R

Conductance Defined

• Conductance is the reciprocal of resistance.

• The units of conductance are called siemens (S)

• The circuit symbol is G

1G

R

i v G i

vG

2p v G 2i

pG

Creating a Circuit Model• A circuit model is usually two or more

circuit elements that are connected.• A circuit model may have active elements

(sources) as well as passive elements (such as resistors).

• By the assumption that electric signal propagation is instantaneous in a circuit, our circuit model has lumped parameters.

Example of a Circuit Model1000 ft AWG 14 Copper Wire

100 WLamp

120 V Battery

120 V

0.25 2.57

2.57

144

Kirchhoff’s Voltage Law

• The sum of the voltage drops around a closed path is zero.

• Example: -120 + V1 + V2 + V3 + V4 = 0

120 V

0.25 2.57

2.57

144

+ V1 - + V2 -

- V4 +

+V3

-

Kirchhoff’s Current Law

• A node is a point where two or more circuit elements are connected together.

• The sum of the currents leaving a node is zero.

I1

I2I3

I4

1 2 3 4 0I I I I

Apply KCL to Example

120 V

0.25 2.57

2.57

144

+ V1 - + V2 -

- V4 +

+V3

-

I s

I1 I1 I2 I2

I3

I3

I4I4

I s

1 2 3 4sI I I I I

Combine KVL, KCL & Ohm’s Law

120 V

0.25 2.57

2.57

144

+ V1 - + V2 -

- V4 +

+V3

-

I s

I1 I1 I2 I2

I3

I3

I4I4

I s

120 0.25 2.57 144 2.57s s s sI I I I

1200.803

149s

VI A

Lamp Voltage & Battery Voltage

3 144 115.67sV I V

(2.57 2 144) 0.803 119.8bV V

120 V

0.25 2.57

2.57

144

+ V1 - + V2 -

- V4 +

+V3

-

I s

I1 I1 I2 I2

I3

I3

I4I4

I s

+

Vb

-

Battery Power and Lamp Power

Loss:

Efficiency:

119.8 0.8033 96.23bP V A W

115.67 0.8033 92.91lP V A W

3.32loss b lP P P W

92.9196.55%

96.23l

b

P

P

1000 ft AWG 14 Copper Wire

100 WLamp

120 V Battery

Using Loops to Write Equations

KVL @Loop a:

KVL @ Loop b:

KVL @ Loop c:

Loop c equation same as a & b combined.

va

R2

vb

R1 R3

+ v2 -

+v1

-

+v3

-a b

c

2 1 0av v v

3 1 0bv v v

2 3 0a bv v v v

Using Nodes to Write Equations

KCL @ Node x:

KCL @ Node y:

KCL @ Node z:KCL @ Node w: <== Redundant

va

R2

vb

R1 R3

+ v2 -

+v1

-

+v3

-

xy z

w

ia

i2 i2 ib ib

iai3

i1

i1

i3

2 1 0bi i i

2 0ai i

3 0bi i 1 3 0ai i i

Combining the Equations• There are 5 circuit elements in the problem.• va and vb are known.• R1, R2 and R3 are known.• v1, v2 and v3 are unknowns.• ia, ib, i1, i2 and i3 are unknowns.• There are 2 loop (KVL) equations.• There are 3 node (KCL) equations.• There are 3 Ohm’s Law equations.• There are 8 unknowns and 8 equations.

Working with Dependent Sources

KVL @ left loop:

KCL @ top right node:

Substitute and solve:

48 V

4

3 i

i+vo

-

i

48 4 3 oV i i

4oi i

3i A 36ov V

Example 1 (1/3)

By KCL:

By Ohm’s Law:

50 V

20 A

25

30 A

50 V10

+ Va - + Vb -+ Vc

-

+Vd

-

Ie

If

Ic Id

20 , 30 , 30 , 10e d f ci A i A i A i A

25 250 , 10 300c c d dV I V V I V

Example 1 (2/3)

By KVL:

Power:

50 V

20 A

25

30 A

50 V10

+ Va - + Vb -+ Vc

-

+Vd

-

Ie

If

Ic Id

300 , 600a bV V V V

300 20 6.0aP V A kW

600 30 18.0bP V A kW

Example 1 (3/3)

50 V

20 A

25

30 A

50 V10

+ Va - + Vb -+ Vc

-

+Vd

-

Ie

If

Ic Id

250 10 2.5cP V A kW

300 30 9.0dP V A kW

50 20 1.0eP V A kW 50 30 1.5fP V A kW

Example 2 (1/4)

Find Source Current, I, and Resistance, R.

1

84 V4

12

8

12 R

8

3 A

I

Example 2 (2/4)

Ohm’s Law: 36 V KVL: 48 V Ohm’s Law: 6 A

1

84 V4

12

8

12 R

8

3 A

I+

36 V-

+48 V

-

6 A

Example 2 (3/4)

KCL: 3 A Ohm’s Law: 12 V KVL: 60 V

1

84 V4

12

8

12 R

8

3 A

I+

36 V-

+48 V

-

6 A

3 A -12 V+

+ 60 V-

Example 2 (4/4)

Ohm’s Law: 3 A KCL: 6 A

Ohm’s Law: R=3 W KCL: I=9 A

KVL: 24 V

1

84 V4

12

8

12 R

8

3 A

I+

36 V-

+48 V

-

6 A

3 A -12 V+

+ 60 V-

+ 24 V -

3 A

6 A