INC 111 Basic Circuit Analysis Week 5 Thevenin’s Theorem.

INC 111 Basic Circuit Analysis

Week 5

Thevenin’s Theorem

Special Techniques

• Superposition Theorem

• Thevenin’s Theorem

• Norton’s Theorem

• Source Transformation

Linearity Characteristic

RL

6Ω

42V

4Ω

10V

IL+

VRL

-

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

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

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

RL

6Ω

42V

4Ω

10V

RL

RTH

VTH

Equivalent circuit

RL

6Ω

42V

4Ω

10V

RL

RTH

VTH

I

V

VOC

ISC

Voc = Voltage open-circuitIsc = Current short-circuitRth = Rth equivalent

Slope =thR

1

RLchangevalue

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.)

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)

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.

Example

3Ω

2Ω

10V

10Ω

RL

2Ω

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

3Ω

2Ω

10V

10Ω

2Ω

0V

10V

0V 0V

6V 6V

VVTH 61032

3

Find VTH

3Ω

2Ω

10V

10Ω

2Ω

3Ω

2Ω 10Ω

2Ω

Short voltage source

RTH

2.13

232

3210

23||210THR

Find RTH

13.2Ω

6V RL

Thevenin’s equivalent circuit

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

6

Example

3Ω

2Ω 10Ω

RL

2Ω

1A

Find Thevenin’s equivalent circuit

Find VTH

3Ω

2Ω 10Ω

2Ω

1A

0V

5V

0V 0V

3V 3V

VVTH 331

3Ω

2Ω 10Ω

2Ω

Open circuitcurrent source

RTH

15

2310THR

Find RTH

3Ω

2Ω 10Ω

2Ω

1A

15Ω

3V RL


R3=4K

R2=8K

R1=2K

R4=1K

RL=1K10V

+ -

Example: Bridge circuit

Find Thevenin’s equivalent circuit

Find VTH

R3=4K

R2=8K

R1=2K

R4=1K

10V

0V

10V

8V 2V

VTH = 8-2 = 6V

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

R3=4K

R2=8K

R1=2K

R4=1K

KKK

KKKKRTH4.28.06.1

1||48||2

2.4K

6V RL


Special Techniques

• Superposition Theorem

• Thevenin’s Theorem

• Norton’s Theorem

• Source Transformation

I

VVOC

ISC

For any point in linear circuit

Thevenin’s Equivalent Circuit

RL

RTH

VTH

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.)

Example

3Ω

2Ω

10V

10Ω

RL

2Ω

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

3Ω

2Ω

10V

10Ω

2Ω

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

3Ω

2Ω

10V

10Ω

2Ω

3Ω

2Ω 10Ω

2Ω

Short voltage source

RTH

2.13

232

3210

23||210THR

Find Rn

Norton’s equivalent circuit

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

2.13

RL13.20.45

Relationship BetweenThevenin’s and Norton’s Circuit

THNTH

NTH

RIV

RR

I

VVOC

ISC

Slope = - 1/Rth

RL13.20.45


13.2

6V RL


Same R value

2.1345.06 THNTH

NTH

RIV

RR

Example

3Ω

2Ω 10Ω

RL

2Ω

1A

Find Norton’s equivalent circuit

Find In

3Ω

2Ω 10Ω

2Ω

1A Isc

Current divider AISC 2.01123

3

3Ω

2Ω 10Ω

2Ω

Open circuitcurrent source

RTH

15

2310THR

Find RTH

3Ω

2Ω 10Ω

2Ω

1A


RL150.2

15

3V RL

Thevenin’s equivalent circuitNorton’s equivalent circuit

RL150.2

0.2 x 15 = 3

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

Example

1V 4K

2K

80

250

RL+

Vx- -

+ 100Vx

+

-

Find Thevenin and Norton’s equivalent circuit

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

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

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

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

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

INC 111 Basic Circuit Analysis Week 5 Thevenin’s Theorem.

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