Combining Voltage Sources PDF

Preview

Summary

Teacher: Jones
Notes about Combining Voltage Sources...

Description

EET 300 || Chapter 1(D) Lesson Notes

Combining Sources

1 OF 5

9/6/2011

Combining Voltage Sources V2

- + V1

+ -

+

KVL Loop

Question:

+ -

Veq

By KVL, Veq

0  V1  V2  Veq

Veq  V1  V2

-

Parallel voltage sources are not allowed! Why?

Let’s look at two parallel voltage sources and apply a KVL loop to them.

8v

+ ‐

KVL

+ ‐

6v

0   8v  6v That's why parallel voltage sources are NOT ALLOWED!

Example: Combine the voltage sources into a single source. 5v

+ 10v

+ -

KVL Loop

+ Veq -

Special note should be taken of the polarities!

+ -

Veq 5v

0   10v  5v  Veq

Veq  10v  5v  5v

EET 300 || Chapter 1(D) Lesson Notes

Combining Sources

2 OF 5

9/6/2011

Combining Current Sources I eq

I1

By KCL, Ieq

I2

0  I 1  I 2  Ieq

Ieq  I1  I2

Current Sources in series are not allowed! This should be especially obvious. Consider the fact that current in a series circuit is the same everywhere. However, if you had two current sources of different values in the series circuit then that would cause the all the same rule to be broken. Thus, it “DOES NOT WORK!” and “IS NOT ALLOWED”. Example: Combine the two current sources in the circuit shown into a single source.

Ieq

I1 3 mA

0   I1  I2  Ieq

I2 5mA

I eq

8mA

By KCL

Ieq  3mA  5mA  8mA

EET 300 || Chapter 1(D) Lesson Notes

Combining Sources

3 OF 5

9/6/2011

Dependent Source Models A dependent source has an output which depends on some other voltage or current in the circuit. The dependent source is denoted by a diamond as can be seen in the four models from EET 210 below.

VCVS + Vi -

+ -

VCIS + Vi -

A vi

Basic amp model EET 210,220,330 A = voltage gain (Unitless)

gm vi

FET and Vaccum tube model g m = transconductance (1/  )

R m ii

R m = transresistance

ICVS ii

+ -

ICIS ii

 ii

BJT Model

 = Current gain

EET 300 || Chapter 1(D) Lesson Notes

Combining Sources

4 OF 5

9/6/2011

Dependent Source Example: Determine: The voltage across the A: dependent source. The Voltage, Vo B: 3A C: The power absorbed/supplied by the dependent source. The power absorbed/supplied by D: the independent source. Check the power balance equation. E:

8 + VR + Vo -

+ -

KVL

2VR

Solution:

KVL :

0  -Vo  VR  2VR

Vo  3VR But

VR  3A(8 )  24 v

so,

Vo  72v

 B

The Voltage across the Dependent Source Vdep  2VR

8  + 72v -

KVL

 48v  A

PInd  3A (72v )  216W PR  3A (24v )  72W Pdep  3(48)  144W

+ 24v -

3A

 2(24)

+ -

2VR

(Delivered )  D (Absorbed ) (Absorbed ) C

Power Balance Equation Power Delivered = Pow er Absorbed ? PInd  PR  Pdep 216W  72W  144W 216W  216W

The Multisim Proof of this work follows:

BALANCE D!

E

EET 300 || Chapter 1(D) Lesson Notes

Combining Sources

5 OF 5

9/6/2011

Let’s take a moment and consider the meaning of voltage and current polarity signs in a circuit. Current Arrows: A current arrow in a circuit represents an imaginary ammeter in the circuit. The direction that the arrow is pointing indicates the manner in which the imaginary ammeter is hooked up. It has NOTHING to do with actual direction of the current. For instance, observe the arrow in the figure shown below (I T ). The observed current in the circuit is 3A going in a clockwise direction due to the current source in a series circuit. However, the drawn current arrow indicates that the ammeter is hooked up in such a way that it will read -3A. Therefore, the current reading you would document from your CALCULATIONS would be IT = -3A.

IT The same thing applies to voltage polarities shown on + 3A 72v circuit. The VR variable shown to the right represents a voltmeter hooked up with the positive terminal connected to the left of the resistor and the negative terminal connected to the right of the resistor. It will read +24V.

8  + VR -

KVL

+ -

2VR...