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INC 112 Basic Circuit Analysis Week 5 Thevenin’s Theorem

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INC 112 Basic Circuit Analysis. Week 5 Thevenin’s Theorem. Special Techniques. Superposition Theorem Thevenin’s Theorem Norton’s Theorem Source Transformation. Linearity Characteristic. If R L change its value , how will it effect the current and voltage across it?. I. - PowerPoint PPT Presentation

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Page 1: INC 112 Basic Circuit Analysis

INC 112 Basic Circuit Analysis

Week 5

Thevenin’s Theorem

Page 2: INC 112 Basic Circuit Analysis

Special Techniques

• Superposition Theorem

• Thevenin’s Theorem

• Norton’s Theorem

• Source Transformation

Page 3: INC 112 Basic Circuit Analysis

Linearity Characteristic

RL

42V

10V

IL+

VRL

-

If RL change its value , how will it effect the current and voltage across it?

Page 4: INC 112 Basic Circuit Analysis

I

VVOC

ISC

For any circuit constructed from only linear components

Not just RL, all resistors have this property.

Voc = Voltage open-circuit Isc = Current short-circuit

Page 5: INC 112 Basic Circuit Analysis

Thevenin’s TheoremWhen we are interested in current and voltage across RL, we can simplify other parts in the circuit.

RL

42V

10V

RL

RTH

VTH

Equivalent circuit

Page 6: INC 112 Basic Circuit Analysis

RL

42V

10V

RL

RTH

VTH

I

V

VOC

ISC

Voc = Voltage open-circuitIsc = Current short-circuitR = R equivalent

Slope = 1/R

Page 7: INC 112 Basic Circuit Analysis

Thevenin’s Equivalent Circuit

RL

RTH

VTH

Thevenin’sequivalentcircuit

VTH = Voc (by removing RL and find the voltage difference between 2 pins)

RTH (by looking into the opened connections that we remove RL, see how much resistance from the connections. If we see a voltage source, we short circuit. If we see a current source, we open circuit.)

Page 8: INC 112 Basic Circuit Analysis

Why do we needequivalent circuit?

• To analyze a circuit with several values of RL

• For circuit simplification (source transformation)

• To find RL that gives maximum power (maximum power transfer theorem)

Page 9: INC 112 Basic Circuit Analysis

Procedure

1. Remove RL from the circuit

2. Find voltage difference of the 2 opened connections. Let it equal VTH.

3. From step 2 find RTH by3.1 short-circuit voltage sources3.2 open-circuit current sources3.3 Look into the 2 opened connections. Find equivalent resistance.

Page 10: INC 112 Basic Circuit Analysis

Example

10V

10Ω

RL

Find Thevenin’s equivalent circuitand find the current that passes through RL when RL = 1Ω

Page 11: INC 112 Basic Circuit Analysis

10V

10Ω

0V

10V

0V 0V

6V 6V

VVTH 61032

3

Find VTH

Page 12: INC 112 Basic Circuit Analysis

10V

10Ω

2Ω 10Ω

Short voltage source

RTH

2.13

232

3210

23||210THR

Find RTH

Page 13: INC 112 Basic Circuit Analysis

13.2Ω

6V RL

Thevenin’s equivalent circuit

If RL = 1Ω, the current is A423.012.13

6

Page 14: INC 112 Basic Circuit Analysis

Example

2Ω 10Ω

RL

1A

Find Thevenin’s equivalent circuit

Page 15: INC 112 Basic Circuit Analysis

Find VTH

2Ω 10Ω

1A

0V

5V

0V 0V

3V 3V

VVTH 331

Page 16: INC 112 Basic Circuit Analysis

2Ω 10Ω

Open circuitcurrent source

RTH

15

2310THR

Find RTH

2Ω 10Ω

1A

Page 17: INC 112 Basic Circuit Analysis

15Ω

3V RL

Thevenin’s equivalent circuit

Page 18: INC 112 Basic Circuit Analysis

R3=4K

R2=8K

R1=2K

R4=2K

RL=1K10V

+ -

Example: Bridge circuit

Find Thevenin’s equivalent circuit

Page 19: INC 112 Basic Circuit Analysis

Find VTH

R3=4K

R2=8K

R1=2K

R4=1K

10V

0V

10V

8V 2V

VTH = 8-2 = 6V

Page 20: INC 112 Basic Circuit Analysis

Find RTH

R3=4K

R2=8K

R1=2K

R4=1K

RTH

R3=4K

R2=8K

R1=2K

R4=1K

R3=4K

R2=8K

R1=2K

R4=1K

Page 21: INC 112 Basic Circuit Analysis

R3=4K

R2=8K

R1=2K

R4=1K

KKK

KKKKRTH4.28.06.1

1||48||2

Page 22: INC 112 Basic Circuit Analysis

2.4K

6V RL

Thevenin’s equivalent circuit

Page 23: INC 112 Basic Circuit Analysis

Special Techniques

• Superposition Theorem

• Thevenin’s Theorem

• Norton’s Theorem

• Source Transformation

Page 24: INC 112 Basic Circuit Analysis

I

VVOC

ISC

For any point in linear circuit

Page 25: INC 112 Basic Circuit Analysis

Thevenin’s Equivalent Circuit

RL

RTH

VTH

Page 26: INC 112 Basic Circuit Analysis

Norton’s Equivalent Circuit

RLRNIN

In= Isc from replacing RL with an electric wire (resistance = 0) and find the currentRn = RTH (by looking into the opened connections that we remove RL, see how much resistance from the connections. If we see a voltage source, we short circuit. If we see a current source, we open circuit.)

Page 27: INC 112 Basic Circuit Analysis

Example

10V

10Ω

RL

Find Norton’s equivalent circuitand find the current that passes through RL when RL = 1Ω

Page 28: INC 112 Basic Circuit Analysis

10V

10Ω

Isc

Find In

Find R total

Find I total

Current divider

4.4123

1232)210(||32

AR

VI 27.2

4.4

10

AISC 45.027.2123

3

Page 29: INC 112 Basic Circuit Analysis

10V

10Ω

2Ω 10Ω

Short voltage source

RTH

2.13

232

3210

23||210THR

Find Rn

Page 30: INC 112 Basic Circuit Analysis

Norton’s equivalent circuit

If RL = 1Ω, the current is A418.045.012.13

2.13

RL13.20.45

Page 31: INC 112 Basic Circuit Analysis

Relationship BetweenThevenin’s and Norton’s Circuit

THNTH

NTH

RIV

RR

I

VVOC

ISC

Slope = - 1/Rth

Page 32: INC 112 Basic Circuit Analysis

RL13.20.45

Norton’s equivalent circuit

13.2

6V RL

Thevenin’s equivalent circuit

Same R value

2.1345.06 THNTH

NTH

RIV

RR

Page 33: INC 112 Basic Circuit Analysis

Example

2Ω 10Ω

RL

1A

Find Norton’s equivalent circuit

Page 34: INC 112 Basic Circuit Analysis

Find In

2Ω 10Ω

1A Isc

Current divider AISC 2.01123

3

Page 35: INC 112 Basic Circuit Analysis

2Ω 10Ω

Open circuitcurrent source

RTH

15

2310THR

Find RTH

2Ω 10Ω

1A

Page 36: INC 112 Basic Circuit Analysis

Norton’s equivalent circuit

RL150.2

Page 37: INC 112 Basic Circuit Analysis

15

3V RL

Thevenin’s equivalent circuitNorton’s equivalent circuit

RL150.2

0.2 x 15 = 3

Page 38: INC 112 Basic Circuit Analysis

Equivalent Circuits withDependent Sources

We cannot find Rth in circuits with dependent sources usingthe total resistance method.

But we can use

SC

OCTH I

VR

Page 39: INC 112 Basic Circuit Analysis

Example

1V 4K

2K

80

250

RL+

Vx- -

+ 100Vx

+

-

Find Thevenin and Norton’s equivalent circuit

Page 40: INC 112 Basic Circuit Analysis

1V 4K

2K

80

250

+Vx- -

+ 100Vx

+

-

Find Voc

I1 I2

12400014250

0)21(400012501

II

IIIKVLloop1

024060801404000

)21(4000

010028022000)12(4000

II

IIVx

VxIIIIKVLloop2

Page 41: INC 112 Basic Circuit Analysis

1V 4K

2K

80

250

+Vx- -

+ 100Vx

+

-

I1 I2

Solve equations

I1 = 3.697mA I2 = 3.678mA

V

mA

IIIVxIVOC

3.7

)678.3697.3(400000)678.3(80

)21(400000280100280

Page 42: INC 112 Basic Circuit Analysis

1V 4K

2K

80

250

+Vx- -

+ 100Vx

Isc

Find Isc

I1 I2 I3

12400014250

0)21(400012501

II

IIIKVLloop1

038024060801404000

)21(4000

0100)32(8022000)12(4000

III

IIVx

VxIIIIIKVLloop2

KVLloop3

038024000801400000

0100)23(80

III

VxII

Page 43: INC 112 Basic Circuit Analysis

1V 4K

2K

80

250

+Vx- -

+ 100Vx

Isc

Find Isc

I1 I2 I3

I1 = 0.632mAI2 = 0.421mAI3 = -1.052 A

Isc = I3 = -1.052 A

Page 44: INC 112 Basic Circuit Analysis

94.6052.1

28.7

SC

OCTH I

VR

6.94

-7.28V RL

Thevenin’s equivalent circuit Norton’s equivalent circuit

RL6.94-1.052