EE 201 node-voltage method – 1

Superposition• Equivalent resistance

• Voltage / current dividers

• Source transformations

• Node voltages

• Mesh currents

• Superposition

In a circuit having more than one independent source, we can consider the effects of the sources one at a time.

EE 201 superposition – 2

�2va + 6vb = 2 V8va � 2vb = 36 V

Solving gives: va = 5 V, vb = 2 V

8v�a � 2v�

b = 0�2v�

a + 6v�b = 2 V

Solving gives: v’a = 0.09091 V, vb = 0.363637 V

�2v��a + 6v��

b = 0

8v��a � 2v��

b = 36 VSolving gives: v��

a = 4.909091 V, v��b = 1.636364 V

v�a + v��

a = 5.0 V = vav�b + v��

b = 2.0 V = vb

A math problem:

Mathematically, we can solve the simultaneous equations a bit at a time.

This means that we can solve a circuit a bit at a time.

EE 201 superposition – 3

The superposition method

In a circuit having more than one independent source, we can consider the effects of the sources one at a time.

If a circuit has n independent sources, then we will have to solve n separate circuits. It this easier? Perhaps. The resulting “partial” circuits will have one source and some resistors. We might be able to solve the partial circuits using the short-cut methods we saw earlier – each partial circuit may go very fast.

As we consider the effect of each source by itself, we must “turn off” (de-activate) all of the other sources – set their values to zero.

VS0 V

IS0 A

same as

same as

short circuit

open circuit

Replace voltage sources with shorts. Replace current sources with opens.

EE 201 superposition – 4

Example of the superposition methodConsider the 2-source, 2-resistor circuit one more time. Let’s us use superposition to find the voltage vR2.

1. Turn off one of the sources – the order doesn’t matter. So start by de-activating the current source. Set the value of IS to zero, which has the same effect as replacing IS with an open circuit.

Y�5� =

5�5� + 5�

96

=�ż

��ż + �ż (��9) = �.��9

+–VS

R1

R2 IS–

+vR2

10 V 1 A

10 !

5 !

10 V

10 !

+–VS

R1

R2–

+Y�5�

5 !

EE 201 superposition – 5

2. Go back to the original circuit and turn off the other source – set VS to zero, which is the same as replacing it with a short circuit.

Note that shorting VS causes R1 to be in parallel with R2.

Y��5� = ,6 (5��5�)

= (�$) (�ż���ż) = �.��9

3. The complete answer is the sum (superposition) of the two partial answers.

Y5� = Y�5� + Y��

5� = �.��9+ �.��9 = �.��9

R1

R2 IS–

+Y��5�

EE 201 superposition – 6

Summary of the superposition method1. Identify all of the independent sources in the circuit.

2. Choose one source that will remain active. De-activate all of the others. (Remove current sources, leaving open circuits. Replace voltage sources with short circuits.)

3. Using whatever techniques are appropriate, solve for the desired quantity (current or voltage) in the circuit. This will be a “partial” result, due only to the one active source in the circuit.

4. Return to the original circuit. Choose a different source to remain active and de-activate all of the others.

5. Solve again for the desired quantity, which will be a second partial result.

6. Continue in this manner, working sequentially through each of the sources in the circuit, finding a partial result for each.

7. Add together all of the partial results to obtain the total result corresponding to all of the sources working simultaneously in the circuit.

EE 201 superposition – 7

Cautions in using superposition

1. Superposition only works with linear circuits. (Linear circuits contain only sources, resistors, capacitors, inductors, linear amplifiers, etc.) Most electronic devices (diodes and transistors) are non-linear, so superposition will not be applicable.

2. Because the method relies on linearity, you cannot add powers directly using the superposition method. (Power goes as v2 or i2 – it is not linear.) Use superposition to find the total current or voltage and then calculate power from that result.

3. When finding the partial solutions, be sure to maintain the same voltage polarity and current direction in each case. For example, one source may induce the current in a particular resistor to flow in one direction while another source causes to a current flowing in the opposite direction. You must keep the proper signs when adding to the partial results to obtain the correct total result.

EE 201 superposition – 8

Example 2

+–

+–

VS1 VS2

R1

R2

R3

R4

R5

In the circuit, find vR2. With two sources, there will be two partial solutions.

50 V

10 ! 50 !

20 !

40 ! 150 !

175 V

Consider VS1 first, so de-active VS2. We see that the series combination of R4 and R5 is in parallel with R2. So the equivalent resistance there is

+–

VS1

R1

R2

R3

R4

R5–

+v�R2

R245 = R2�� (R4 + R5) = (20 Ω) � (50 Ω+ 150 Ω) = 18.2 Ω

Then, using a voltage divider:

v�R2 =

R245R1 + R245 + R3

VS1 =(18.2 Ω)

(10 Ω) + (18.2 Ω) + (40 Ω)(50 V) = 13.34 V

EE 201 superposition – 9

Next, consider VS2 , with VS1 de-activated. This partial circuit is very similar to the first case, but from the other side. We see that the series combination of R1 and R3 is in parallel with R2. So the equivalent resistance, seen looking from the right is

+–VS2

R1

R2

R3

R4

R5–

+v��R2

R123 = R2�� (R1 + R3) = (20 Ω) � (10 Ω+ 40 Ω) = 14.3 Ω

Using the voltage — from the right this time:

v��R2 =

R123R4 + R123 + R5

VS2 =(14.3 Ω)

(50 Ω) + (14.3 Ω) + (150 Ω)(175 V) = 11.66 V

Then the total voltage is the sum of the two partial voltages:

vR2 = v�R2 + v��

R2 = 13.34 V+ 11.66 V = 25.0 V

as we have seen using other techniques.

EE 201 superposition – 10

Example 3

In the circuit, find iR3. With three sources, there will be three partial solutions.

1. Start with VS. De-activate the two current sources. Find in the partial circuit, using whatever short-cut methods you like. One approach: use a source transformation and then a current divider.

L�5� =�

5�+5��5� + �

5� + �5�+5�

�965�

�

10 V

1 k!

2 k!

10 mA

5 mA

1 k!

2 k!

=�

�Nż�

�Nż + ��Nż + �

�Nż(��P$) = �.���P$

+–

VS

R1

R2

R3

R4

IS1

IS2iR3

+–

VS

R1

R2

R3

R4L�5�i�R3

R1 R2

R3

R4L�5�965�

10 V

1 k!

2 k!

1 k!

2 k!

2 k!

1 k!

2 k!1 k!10 mA

EE 201 superposition – 11

2. Go back to the original circuit and de-activate VS and IS2, keeping IS1. Note that with VS replaced by a short, R1 is in parallel with R2.

Examining the partial circuit carefully, we see that it is essentially a current divider. The source current splits between two paths – one with R3 and another that is a combination of R1, R2, and R4.

= �Nż��Nż + �Nż = �.��Nż

=�

�Nż�

�Nż + ��.��Nż

(��P$) = �.��P$

5�� = 5��5� + 5�

L��5� =�5�

�5� + �

5��,6�

R1

R2

R3

R4

IS1

L��5�1 k!

2 k!

10 mA

1 k!

2 k!

EE 201 superposition – 12

3. Now we need the final partial circuit. Go back to the original circuit and de-activate VS and IS1, keeping IS2. Again, R1 is in parallel with R2.

Again, we can handle the calculation with a current divider. The source current splits between two paths – one with R4 and another that is a combination of R1, R2, and R3. Also, note carefully the direction – will be negative!

= �Nż��Nż + �Nż = �.��Nż

= ��

�.��Nż�

�.��Nż + ��Nż

(�P$) = ��.��P$

= �.���P$+ �.��P$� �.��P$ = �.��P$L5� = L�5� + L��5� + L���5�

4. To complete the calculation, add the three partial results to get the total.

5�� = 5��5� + 5�

L���5� = ��

5���

5�� + �5�

,6�

R1

R2

R3

R4 IS2L���5�1 k!

2 k! 5 mA

1 k!

2 k!

i���R3

EE 201 superposition – 13

Example 4

Find vR2.

Looks nasty – 6 nodes, 4 meshes. Either NV or MC will be tough. Might be able to do a bunch of source transformations. Since there are only three sources, let’s try superposition.

100 V

75 !

75 !

50 !

50 !

75 !

50 !

75 V5 A

1. Keep VS1 and de-activate IS and VS2. Shorting VS2

puts R6 in parallel with R5.

Use a voltage divider between R1 and all the rest.

v0R2 =

R26

R1 + R26VS1 =

��.�ż��ż + ��.�ż (���9) = ��.�9

5�� = 5�� (5� + 5� + 5��5�) = ��.�ż

+–

+–

VS1

R1 R3 R6

R4

R2 R5 VS2IS–

+vR2

+–

VS1

R1 R3 R6

R4

R2 R5–

+Y�5�

100 V

75 !

75 !

50 !

50 !

75 !

50 !

EE 201 superposition – 14

2. Go back to the original circuit. Keep VS2 and de-activate IS and VS1. Shorting VS1 puts R1 in parallel with R2.

Use voltage dividers twice: first to find the voltage across R5, then again to find voltage across R2.

Y��5� =

5��5�5� + 5� + 5��5�

Y��5�

Y��5� =

5��5� + 5��

96� =��.��ż

��ż + ��.��ż (��9) = ��.��9

=��.�ż

��ż + ��ż + ��.�ż (��.��9) = �.��9

+–

R1 R3 R6

R4

R2 R5 VS2–

+Y��5�

–

+Y��5�

R51 = R5���� �R3 + R4 + R1

����R2�

= 36.67 Ω

R1 R3

R4

R2–

+Y��5�

–

+Y��5� = ��.�� 9

75 !

75 !

50 !

50 !

75 !

50 !

75 V

EE 201 superposition – 15

3. Finally, keep IS and

de-activate VS1 and VS2 in the original circuit.

Source current divides to the left and right according to the equivalent resistances on either side. Use a current divider to find the current to the left.

L/ =�

5���5��

+ �5��

,6 =�

���.�ż�

���.�ż + ���ż

(�$) = �.��$

4. Add the three partial results to get the total.

Y5� = Y�5� + Y��

5� + Y���5� = ��.�9+ �.��9+ ��.�9 = ��.�9

R1 R3 R6

R4

R2 R5IS–

+Y���5�

iL

R41 = R4 + R3 + R1����R2 = 137.5 ٠R56 = R5

����R6 = 30 Ω

v���R2 = iL

�R1

����R2�

= (0.90 A)�75 Ω

����75 �

= 33.6 V

75 !

75 !

50 !

50 !

75 !

50 !

5 A