CIRCUIT THEOREMS

Anuroop Gaddam

An ideal voltage source plots a vertical line on the VI characteristic as shown for the ideal 6.0 V source.

Actual voltage sources include the internal source resistance, which can drop a small voltage under load. The characteristic of a non-ideal source is not vertical.

Cur

rent

(A)

Voltage (V)

0 0

1

2

2

3

4

4 6

5

8 10

Voltage sources

A practical voltage source is drawn as an ideal source in series with the source resistance. When the internal resistance is zero, the source reduces to an ideal one.

RS

VS+

Voltage sources

If the source resistance of a 5.0 V power supply is 0.5 W, what is the voltage across a 68 W load?

Use the voltage-divider equation:

LL S

L S

68 5 V = 4.96 V

68 0.5

RV V

R R

RS

RL

VS

VOUT

+5.0 V

0.5

68

An ideal current source plots a horizontal line on the VI characteristic as shown for the ideal 4.0 mA source.

Current sources

Practical current sources have internal source resistance, which takes some of the current. The characteristic of a practical source is not horizontal.

Cur

rent

(A)

Voltage (V)

0 0

1

2

2

3

4

4 6

5

8 10

A practical current source is drawn as an ideal source with a parallel source resistance. When the source resistance is infinite, the current source is ideal.

Current sources

RSIS

If the source resistance of a 10 mA current source is 4.7 kW, what is the voltage across a 100 W load?

Use the current-divider equation:

SL S

L S

4.7 k10 mA = 9.8 mA

100 4.7 k

RI I

R R

RSIS RL

100 4.7 k10 mA

Source conversionsAny voltage source with an internal

resistance can be converted to an equivalent current source and vice-versa by applying Ohm’s law to the source. The source resistance, RS, is the same for both.

To convert a voltage source to a current source, S

SS

VI

R

S S SV I RTo convert a current source to a voltage source,

Superposition theorem

The superposition theorem is a way to determine currents and voltages in a linear circuit that has multiple sources by taking one source at a time and algebraically summing the results. What does the ammeter read for I2? (See next slide for the method and the answer).

+-

-

+

-

+R 1 R 3

R 2

I 2

V S 2V S 1

1 2 V

2 .7 k 6 .8 k

6 .8 k

1 8 V

6.10 kW

What does the ammeter read for I2?

1.97 mA 0.98 mA

8.73 kW 2.06 mA

+-

-

+

-

+

R 1 R 3

R 2

I 2

V S 2V S 1

1 2 V

2 .7 k 6 .8 k

6 .8 k

1 8 V

0.58 mA

1.56 mA

1.56 mA

Source 1: RT(S1)= I1= I2= Source 2: RT(S2)= I3= I2= Both sources I2=

Set up a table of pertinent information and solve for each quantity listed:

The total current is the algebraic sum.

Thevenin’s theorem states that any two-terminal, resistive circuit can be replaced with a simple equivalent circuit when viewed from two output terminals. The equivalent circuit is:

Thevenin’s theorem

V T H

R T H

V T H

R T H

VTH is defined as

Thevenin’s theorem

RTH is defined as

the open circuit voltage between the two output terminals of a circuit.

the total resistance appearing between the two output terminals when all sources have been replaced by their internal resistances.

Thevenin’s theorem

R

R

1

R 2R 2 L

V SV S

1 2 V1 0 k

6 8 k 2 7 k

Output terminals

What is the Thevenin voltage for the circuit? 8.76 V

What is the Thevenin resistance for the circuit? 7.30 kW

Remember, the load resistor has no affect on the Thevenin parameters.

Thevenin’s theorem

Thevenin’s theorem is useful for solving the Wheatstone bridge. One way to Thevenize the bridge is to create two Thevenin circuits - from A to ground and from B to ground.

The resistance between point A and ground is R1||R3 and the resistance from B to ground is R2||R4. The voltage on each side of the bridge is found using the voltage divider rule. R3 R4

R2

RL

R1VS

-

+A B

Thevenin’s theorem

For the bridge shown, R1||R3 = and R2||R4 = . The voltage from A to ground (with no load) is and from B to ground (with no load) is .

The Thevenin circuits for each of the bridge are shown on the following slide.

165 W179 W

7.5 V 6.87 V

R3 R4

R2

RL

R1VS

-

+A B

330 390

330 330

+15 V150

Thevenin’s theorem

Putting the load on the Thevenin circuits and applying the superposition theorem allows you to calculate the load current.

RLA B

150 VTH VTH

RTH RTH'

'165 179

7.5 V 6.87 V

A B

VTH VTH

RTH RTH'

'165 179

7.5 V 6.87 V

1.27 mAThe load current is:

Norton’s theorem states that any two-terminal, resistive circuit can be replaced with a simple equivalent circuit when viewed from two output terminals. The equivalent circuit is:

Norton’s theorem

RNIN

Norton’s theorem

the output current when the output terminals are shorted.

the total resistance appearing between the two output terminals when all sources have been replaced by their internal resistances.

IN is defined as

RN is defined as

RNIN

Norton’s theorem

Output terminals

What is the Norton current for the circuit? 17.9 mA

What is the Norton resistance for the circuit? 359 W

R2

R1

RL

VS +10 V

560

820 1.0 k

The Norton circuit is shown on the following slide.

Norton’s theorem

RNIN17.9 mA 359 W

The Norton circuit (without the load) is:

Maximum power transfer

The maximum power is transferred from a source to a load when the load resistance is equal to the internal source resistance.

The maximum power transfer theorem assumes the source voltage and resistance are fixed.

RS

RL

VS +

Maximum power transfer

What is the power delivered to the matching load?

The voltage to the load is 5.0 V. The power delivered is

RS

RL

VS + 50

50 10 V

22

LL

5.0 V= 0.5 W

50

VP

R