09-Circuit Theorems Overview Introduction - Henry Louie, PhD

09-Circuit Theorems Text: 4.1 – 4.3, 4.8

ECEGR 210

Electric Circuits I

Overview

• Introduction

• Linearity

• Superposition

• Maximum Power Transfer

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Introduction

• Nodal and mesh analysis can be tedious to apply to large circuits, especially with multiple sources

• Linearity of the circuits considered can be exploited to simplify analysis

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Linearity

• Circuits considered in the this class are linear

• An equation f(x) or system is linear iff

Homogeneity: f(ax) = af(x)

Additivity: f(x + y) = f(x) + f(y)

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Linearity

• Is f(x) = 2x + 1 linear?

• Check homogeneity: f(ax) = af(x)

• Check additivity : f(x + y) = f(x) + f(y)

• No. This is an affine function

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x

f(x) 2ax 1 a(2x 1)

2(x y) 1 2x 1 2y 1

Linearity

• What about Ohm’s Law (V=IR)? Let I be the argument so that f(x) = V(I)


• Check additivity: f(x + y) = f(x) + f(y)

• Yes. A circuit is linear if it consists only of linear elements, linear dependent sources and independent sources

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aIR a(IR)

a b a b(I I )R I R I R

Linearity

• What about P=IV?

Careful. V = IR, so P = I2R


• Check additivity: f(x + y) = f(x) + f(y)

• No. The power dissipated by a resistor is not linear.

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2 2(aI) R aI R

2 2 2

1 2 1 2(I I ) R I R I R

Significance of Linearity

• The mesh currents for the shown circuit can be written as (by inspection):

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1 2 2 1 1 2

2 2 3 4 2 2

R R R i V V

R R R R i V

A (matrix)

1 1 2

2 2

1 1 1 2

2 2

i V V

i V

i V V

i V

A

A

+

-

+

- V1 V2

R1

R2

R3

R4

i1 i2

Solving for i1, i2


• Observing that:

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+

- V1

R1

R2

R3

R4

i1 i2

1 1 1 2 1 1 21

2 2 2

i V V 0 VV 0

i V V0

A A A

+

- V2

R1

R2

R3

R4

i1 i2

We can solve two “easier” circuits, and add the results!


• Find the current out of the dependent voltage source if Vs = 12V, and if Vs = 24V

Should we use nodal or mesh analysis?

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+

-

+

-

2W

6W

4W

8W

4W

3Vx Vs

Vx + -


• since Vs = 12V

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+

-

+

-

2W

6W

4W

8W

4W

3Vx Vs

i1 i2

1 2 s

1 2 x

x 1

0 12i 4i V (mesh 1)

0 4i 16i 3V (mesh 2)

v 2i

Vx + -


Vs = 12V

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+

-

+

-

2W

6W

4W

8W

4W

3Vx Vs

i1 i2

1 2 s

1 2

1 2

1 2

2 s

s2

0 12i 4i V

0 10i 16i

adding...

2i 12i 0

i 6i

via substitution

76i V 0

V 12i 0.158A

76 76

Vx + -


• By linearity: if Vs = 24V (voltage doubled)

i2 = 2x0.158A = 0.316A (current doubled)

• No need to re-solve the circuit

• Behold, the power of linearity!

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+

-

+

-

2W

6W

4W

8W

4W

3Vx Vs

i1 i2

Vx + -

Superposition

• One of the most important circuit analysis benefits of linearity is that superposition can be applied

• Very useful for circuits with multiple independent sources

• Basic idea: analyze the circuit considering only one source at a time, and sum the results once all sources have been considered

Valid for all circuit analysis techniques: Nodal, mesh, Ohm’s Law, etc.

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Superposition

• Simple example with V1 = 10V, V2 = 15V

• Find I

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+

-

+

- V1

5W V2

I

1 2V V

I 1AR

Superposition

• Now solve using superposition

• Consider V1 only

Remove contribution from V2 (set source to 0, a short circuit)

IV1 = 2A

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+

- V1

5W V2 = 0

I

Superposition

• Consider V2 only

Remove contribution from V1 (set source to 0, a short circuit)

IV2 = -3A

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+

- V1 = 0

5W V2 = 15

I

Superposition

• Combining

I = IV2 + IV2 = -1A

Same result as solving the circuit simultaneously

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Superposition

• Analyze circuit with only one independent source at a time

• Independent voltage sources are ignored by replacing them with a short (voltage source where Vs = 0)

• Independent current sources are ignored by replacing them with an open circuit (current source where Is = 0)

• Dependent sources are not ignored

• Sum individual results to analyze circuit

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Example

• Find VR using superposition.

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+

- 6V

8W

4W

+

- VR 3A

Example

• Consider the voltage source first. Set current source to 0A (open circuit)

• By KVL:

12i1 - 6 = 0

i1 = 0.5A

V1 = 2V (VR due to voltage source)

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+

- 6V

8W

4W

+

- VR 3A

+

- 6V

8W

4W +

- V1 i1

Example

• Now consider the current source. Set voltage source to 0V (short circuit)

• By current division:

i2 = (3x8)/(12) = 2A

V2 = 8V (VR due to current source)

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+

- 6V

8W

4W +

- VR 3A

8W

4W +

- V2 3A i2

Example

• Now sum results:

VR = V1 + V2 = 10V

• Note:

iR = i1 + i2 = 2.5A

PR = iR 2R = (i1 + i2 )

2R =31.25W

Any power calculations must be done after all sources considered

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2 2

R 1 2P P P

Example

• Use superposition to find Vx

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20W

4W

+

- Vx 4A

+

- 20V 0.1Vx

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Example

• First consider voltage source.

• Use nodal analysis

Trying to find node voltage

Only one node with unknown voltage

Many elements in parallel

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20W

4W

+

- Vx1

+

- 20V 0.1Vx1

i1 i2

Example

• KCL of node

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x1 1 2

x1 x1x1

x1 x1 x1

x1

i i i

via substitution:

V 20 V0.1V

4 20

V 0.4V 0.2V 4

V 5V

20W

4W

+

- Vx1

+

- 20V 0.1Vx1

i1 i2

ix1

Example

• Now consider current source.

• Rearrange parallel elements for clarity

• Use nodal analysis

Trying to find node voltage

Only one node with unknown voltage

Many elements in parallel

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20W 4W

+

- Vx 0.1Vx 4A

One node

Example

• Now consider current source:

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x2 20 x2

x2 x2x2

x2 x2 x2

x2

4 +0.1V i i

via substitution:

V V4 +0.1V

20 4

4 0.05V 0.25V 0.1V

V 20V

20W 4W

+

- Vx 0.1Vx 4A

One node

Example

• Summing results:

Vx = Vx1 + Vx2 = 25V

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• Consider past example

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1 2 2 1 1 2

2 2 3 4 2 2

R R R i V V

R R R R i V

A (matrix)

+

-

+

- V1 V2

R1

R2

R3

R4

i1 i2


• Assume that we solved for i1 and i2

• If the voltage sources are increased by a factor of x, what happens to i1 and i2?

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new1 21

new22

new1 1 21

new22

V Vix

Vi

V Vix

Vi

A

A

1

2

i

i

new

1 1

new

2 2

i ix

i i

Current also increases by x


• If all the voltages are proportionally increased/decreased, then all current will also proportionally increase/decrease

What happens if all the resistances are proportionally changed?

• Useful if you have already solved the circuit and want to examine how changes will affect the solution

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Maximum Power Transfer

• Often interested in designing a circuit so that the maximum power is delivered (transferred) to a load

• Consider an 11V battery connected to a heater by way of a cable

• What resistance should Rload be to maximize the power it consumes?

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+

- Vs

Rs = 5W

Rload = ?


• Idea: try a small resistance in order to increase current (P = I2R)

Let Rload = 0.5W

• Compute power to the load resistor and power consumed by the cable

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+

- Vs

Rs = 5W

Rload = 0.5W


• More power is consumed by the cable than the load

• Only small portion of applied voltage is across the load

Vload = 1V

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+

- Vs

Rs = 5W

Rload = 0.5W

2 2 2sload load load

load s

2 2

s load

V 11P I R ( ) R ( ) 0.5 2W

R R 0.5 5

11P I R ( ) 5 10W

0.5 5


• New idea: try large resistance so a large voltage appears across the load (P = V2/R)

Let Rload = 50W

• Compute power to the load resistor and power consumed by the cable

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+

- Vs

Rs = 5W

Rload = 0.5W


• More power is consumed by the cable than the load

• Small amount of current flows

I = 0.2A

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+

- Vs

Rs = 5W

Rload = 50W

2 2

load load

2 2

s load

11P I R ( ) 50 2W

5 50

11P I R ( ) 5 10W

5 50


• Trade-off between voltage across load and current through load resistor

• Try Rload = Rs

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• Power to load is increased

Vload = 5.5V

I = 1.1A

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+

- Vs

Rs = 5W

Rload = 5W

2 2

load load

2 2

s load

11P I R ( ) 5 6.05W

5 5

11P I R ( ) 5 6.05W

5 5


• Maximum power transfer occurs when Rload = Rs

• Load and source (with series impedance) are said to be “matched”

• Observations

Voltage across load is one half applied power

Relationship between Rload and Pload must be non-linear

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• Proof:

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2 2sload load load

load s

load

load

2 2

load s s load

2 3

load load s load s

2load load s loads 3

load load s

load s

VP I R ( ) R

R R

dPat maximum power transfer 0

dR

solving

dP V 2V R=

dR (R R ) (R R )

dP R R 2R=V 0

dR (R R )

therefore

R R


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0 5 10 15 20 25 30 35 40 45 500

1

2

3

4

5

6

7

RLoad

(W)

Pow

er

(W)

Rload

too large Rload

too small

09-Circuit Theorems Overview Introduction - Henry Louie, PhD

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Transcript of 09-Circuit Theorems Overview Introduction - Henry Louie, PhD