EE 210:Circuits

Spring 2017

Lecture 7: Thursday, February 2, 2017

Last time (Nodal Analysis)

1. Nodal analysis recipe

2. Nodal analysis examples

Today More Nodal Analysis and Mesh Analysis

1. More nodal analysis examples

2. Modifying the recipe for supernodes

3. Mesh Analysis

4. Source transformations

More Nodal Analysis

1. More nodal analysis examples

2. Modifying the recipe for supernodes

3. Source transformations

4. Mesh Analysis

Nodal Analysis Recipe

1. Label the nodes with unknown voltages and choose a reference node (reference node is

usually the 0V “ground” node)

2. Draw/label the currents at each node (it doesn’t matter which way you orient the arrows)

3. For each node with an unknown voltage:

a) Write a KCL equation: 𝑖𝑖𝑛 = 𝑖𝑜𝑢𝑡

b) Use Ohm’s law to substitute currents for voltages in the KCL equation

4. Solve the set of equations for the voltages

Nodal Analysis Example 4

3Ω

−+ 10Ω

3A

1 2

0Reference node

5Ω

4Ω6V

3Ω

−+ 10Ω

𝐼1

𝐼2

𝐼3 𝐼4

3A

1 2

0Reference node

5Ω

4Ω6V

KCL @ Node 1: 𝐼1 = 𝐼3 + 3

2 − 0.33 3𝑉1 = 0.25𝑉1 + 3

−1 = 0.58 3𝑉1

Ohm’s law: 𝐼1 =6 − 𝑉1

3= 2 − 0.33 3𝑉1

𝐼3 =𝑉1 − 𝑉0

4= 0.25𝑉1

𝐼4 =𝑉2 − 𝑉0

10= 0.1𝑉2

𝐼2 =6 − 𝑉2

5= 1.2 − 0.2𝑉2KCL @ Node 2: 3 + 𝐼2 = 𝐼4

3 + 1.2 − 0.2𝑉2 = 0.1𝑉24.2 = 0.3𝑉2

𝑉1 = −1.71V

𝑉2 = 14V

3Ω

−+ 10Ω

𝐼1

𝐼2

𝐼3 𝐼4

3A

1 2

0Reference node

5Ω

4Ω6V

KCL @ Node 1: 𝐼1 = 𝐼3 + 3

2 − 0.33 3𝑉1 = 0.25𝑉1 + 3

−1 = 0.58 3𝑉1

Ohm’s law: 𝐼1 =6 − 𝑉1

3= 2 − 0.33 3𝑉1

𝐼3 =𝑉1 − 𝑉0

4= 0.25𝑉1

𝐼4 =𝑉2 − 𝑉0

10= 0.1𝑉2

𝐼2 =6 − 𝑉2

5= 1.2 − 0.2𝑉2KCL @ Node 2: 3 + 𝐼2 = 𝐼4

3 + 1.2 − 0.2𝑉2 = 0.1𝑉24.2 = 0.3𝑉2

𝑉1 = −1.71V

𝑉2 = 14V

3Ω

−+ 10Ω

𝐼1

𝐼2

𝐼3 𝐼4

3A

1 2

0Reference node

5Ω

4Ω6V

3Ω

−+ 10Ω

𝐼1

𝐼2

𝐼3 𝐼4

3A

1 2

0Reference node

5Ω

4Ω6V

Ohm’s law: 𝐼1 =6 − 𝑉1

3= 2.57A

𝐼3 =𝑉1 − 𝑉0

4= −0.43A

𝐼4 =𝑉2 − 𝑉0

10= 1.4A

𝐼2 =6 − 𝑉2

5= −1.6A

KCL @ Node 1: 𝐼1 = 𝐼3 + 3

2 − 0.33 3𝑉1 = 0.25𝑉1 + 3

−1 = 0.58 3𝑉1

KCL @ Node 2: 3 + 𝐼2 = 𝐼43 + 1.2 − 0.2𝑉2 = 0.1𝑉24.2 = 0.3𝑉2

𝑉1 = −1.71V

𝑉2 = 14V

−𝟏.𝟔𝐀

3Ω

−+ 10Ω

𝐼1

𝐼2

𝐼3 𝐼4

3A

1 2

0Reference node

5Ω

4Ω6V

Ohm’s law: 𝐼1 =6 − 𝑉1

3= 2.57A

𝐼3 =𝑉1 − 𝑉0

4= −0.43A

𝐼4 =𝑉2 − 𝑉0

10= 1.4A

𝐼2 =6 − 𝑉2

5= −1.6A

𝟏.𝟕𝟏𝐕

𝟏𝟒𝐕𝟔𝐕

𝟎𝐕

−𝟎.𝟒𝟑𝐀 𝟏.𝟒𝐀

𝟐.𝟓𝟕𝐀

KCL @ Node 1: 𝐼1 = 𝐼3 + 3

2 − 0.33 3𝑉1 = 0.25𝑉1 + 3

−1 = 0.58 3𝑉1

KCL @ Node 2: 3 + 𝐼2 = 𝐼43 + 1.2 − 0.2𝑉2 = 0.1𝑉24.2 = 0.3𝑉2

𝑉1 = −1.71V

𝑉2 = 14V

More Nodal Analysis

1. More nodal analysis examples

2. Modifying the recipe for supernodes

3. Mesh Analysis

4. Source transformations

Nodal Analysis Example 5

−+

10Ω

5Ω

−+10V 2Ω 6A

3V

1 2

0Reference node

𝟏𝟎𝐕

𝟎𝐕

𝐼1

𝐼2 𝐼3

𝐼 =?

−+

10Ω

5Ω

−+10V 2Ω 6A

3V

1 2

0Reference node

𝟏𝟎𝐕

𝟎𝐕

𝐼1

𝐼2 𝐼3

𝐼 =?

KCL @ Node 1: 𝐼1 + 𝐼 = 𝐼2

−+

10Ω

5Ω

−+10V 2Ω 6A

3V

1 2

0Reference node

𝟏𝟎𝐕

𝟎𝐕

𝐼1

𝐼2 𝐼3

𝐼 =?

KCL @ Node 1: 𝐼1 + 𝐼 = 𝐼2 Ohm’s law:

𝐼1 =10 − 𝑉1

5= 2 − 0.2𝑉1

𝐼2 =𝑉1 − 𝑉0

10= 0.1𝑉1

0

𝐼3 =𝑉2 − 𝑉0

2= 0.5𝑉2

0

−+

10Ω

5Ω

−+10V 2Ω 6A

3V

1 2

0Reference node

𝟏𝟎𝐕

𝟎𝐕

𝐼1

𝐼2 𝐼3

𝐼 =?

KCL @ Node 1: 𝐼1 + 𝐼 = 𝐼22 − 0.2𝑉1 + 𝐼 = 0.1𝑉1

Ohm’s law:

𝐼1 =10 − 𝑉1

5= 2 − 0.2𝑉1

𝐼2 =𝑉1 − 𝑉0

10= 0.1𝑉1

0

𝐼3 =𝑉2 − 𝑉0

2= 0.5𝑉2

0

−+

10Ω

5Ω

−+10V 2Ω 6A

3V

1 2

0Reference node

𝟏𝟎𝐕

𝟎𝐕

𝐼1

𝐼2 𝐼3

𝐼 =?

KCL @ Node 1: 𝐼1 + 𝐼 = 𝐼22 − 0.2𝑉1 + 𝐼 = 0.1𝑉1

KCL @ Node 2: 6 = 𝐼 + 𝐼36 = 𝐼 + 0.5𝑉2

Ohm’s law:

𝐼1 =10 − 𝑉1

5= 2 − 0.2𝑉1

𝐼2 =𝑉1 − 𝑉0

10= 0.1𝑉1

0

𝐼3 =𝑉2 − 𝑉0

2= 0.5𝑉2

0

−+

10Ω

5Ω

−+10V 2Ω 6A

3V

1 2

0Reference node

𝟏𝟎𝐕

𝟎𝐕

𝐼1

𝐼2 𝐼3

𝐼 =?

KCL @ Node 1: 𝐼1 + 𝐼 = 𝐼22 − 0.2𝑉1 + 𝐼 = 0.1𝑉1

KCL @ Node 2: 6 = 𝐼 + 𝐼36 = 𝐼 + 0.5𝑉2

Ohm’s law:

𝐼1 =10 − 𝑉1

5= 2 − 0.2𝑉1

𝐼2 =𝑉1 − 𝑉0

10= 0.1𝑉1

0

𝐼3 =𝑉2 − 𝑉0

2= 0.5𝑉2

0

Works, but it’s easier to use a supernode

−+

10Ω

5Ω

−+10V 2Ω 6A

3V

1 2

0Reference node

𝟏𝟎𝐕

𝟎𝐕

𝐼1

𝐼2 𝐼3

Ohm’s law:

𝐼1 =10 − 𝑉1

5= 2 − 0.2𝑉1

𝐼2 =𝑉1 − 𝑉0

10= 0.1𝑉1

0

𝐼3 =𝑉2 − 𝑉0

2= 0.5𝑉2

0

𝑉1 = 𝑉2 + 3 [1]

−+

10Ω

5Ω

−+10V 2Ω 6A

3V

1 2

0Reference node

𝟏𝟎𝐕

𝟎𝐕

𝐼1

𝐼2 𝐼3

Ohm’s law:

𝐼1 =10 − 𝑉1

5= 2 − 0.2𝑉1

𝐼2 =𝑉1 − 𝑉0

10= 0.1𝑉1

0

𝐼3 =𝑉2 − 𝑉0

2= 0.5𝑉2

0

𝑉1 = 𝑉2 + 3 [1]

KCL @ Supernode: 𝐼1 + 6 = 𝐼2 + 𝐼3

2 − 0.2𝑉1 + 6 = 0.1𝑉1 + 0.5𝑉2

8 − 0.3𝑉1 = 0.5𝑉2 [2]

−+

10Ω

5Ω

−+10V 2Ω 6A

3V

1 2

0Reference node

𝟏𝟎𝐕

𝟎𝐕

𝐼1

𝐼2 𝐼3

Ohm’s law:

𝐼1 =10 − 𝑉1

5= 2 − 0.2𝑉1

𝐼2 =𝑉1 − 𝑉0

10= 0.1𝑉1

0

𝐼3 =𝑉2 − 𝑉0

2= 0.5𝑉2

0

𝑉1 = 𝑉2 + 3 [1]

KCL @ Supernode: 𝐼1 + 6 = 𝐼2 + 𝐼3

2 − 0.2𝑉1 + 6 = 0.1𝑉1 + 0.5𝑉2

8 − 0.3𝑉1 = 0.5𝑉2 [2]

−+

10Ω

5Ω

−+10V 2Ω 6A

3V

1 2

0Reference node

𝟏𝟎𝐕

𝟎𝐕

𝐼1

𝐼2 𝐼3

Ohm’s law:

𝐼1 =10 − 𝑉1

5= 2 − 0.2𝑉1

𝐼2 =𝑉1 − 𝑉0

10= 0.1𝑉1

0

𝐼3 =𝑉2 − 𝑉0

2= 0.5𝑉2

0

𝑉1 = 𝑉2 + 3 [1]

KCL @ Supernode: 𝐼1 + 6 = 𝐼2 + 𝐼3

2 − 0.2𝑉1 + 6 = 0.1𝑉1 + 0.5𝑉2

8 − 0.3𝑉1 = 0.5𝑉2 [2]

[1] 𝑉1 = 𝑉2 + 3

[2] 8 − 0.3𝑉1 = 0.5𝑉2

Plugging [1] into [2]:

From [1]:

8 − 0.3 𝑉2 + 3 = 0.5𝑉2

7.1 = 0.8𝑉2

8 − 0.3𝑉2 − 0.9 = 0.5𝑉2

𝑉2 = 8.875V

𝑉1 = 𝑉2 + 3

𝑉1 = 11.875V

𝐼1 = 2 − 0.2𝑉1 = −0.375A

𝐼2 = 0.1𝑉1 = 1.1875A

𝐼3 = 0.5𝑉2 = 4.4375A

From Ohm’s law equations:

−+

10Ω

5Ω

−+10V 2Ω 6A

3V

1 2

0Reference node

𝟏𝟎𝐕

𝟎𝐕

𝐼1

𝐼2 𝐼3

𝟏𝟏.𝟖𝟕𝟓𝐕 𝟖.𝟖𝟕𝟓𝐕−𝟎.𝟑𝟕𝟓𝐀

𝟏.𝟏𝟖𝟕𝟓𝐀 𝟒.𝟒𝟑𝟕𝟓𝐀

𝟏.𝟓𝟔𝟐𝟓𝐀

𝑉2 = 8.875V

𝑉1 = 11.875V

KCL @ Supernode: 𝐼1 + 6 = 𝐼2 + 𝐼3

2 − 0.2𝑉1 + 6 = 0.1𝑉1 + 0.5𝑉2

8 − 0.3𝑉1 = 0.5𝑉2 [2]

Ohm’s law:

𝐼1 =10 − 𝑉1

5= 2 − 0.2𝑉1

𝐼2 =𝑉1 − 𝑉0

10= 0.1𝑉1

0

𝐼3 =𝑉2 − 𝑉0

2= 0.5𝑉2

0

−0.375A

1.1875A

4.4375A

Mesh Analysis

1. More nodal analysis examples

2. Modifying the recipe for supernodes

3. Source transformations

4. Mesh Analysis

Source transformations allow you to switch between a voltage source in series with a resistor and a current sourcein parallel with the same resistor.

As demonstrated in the following example, this technique allows you to significantly simplify circuits.

Goal: Find 𝑖0

Now combine 10V and 5V sources (can do this because they are in series)...

Now use the following source transformation to get everything in parallel...

7.5A

Now combine 7.5A and 3A sources (can do this because they are current sources in parallel), and combine the 2Ω and 5Ω resistors in parallel...

At this point, just use current division to find that the current 𝑖0 = 1.78A.

10.5A

7.5A

2Ω ∥ 5Ω =10

7Ω

Mesh Analysis

1. More nodal analysis examples

2. Modifying the recipe for supernodes

3. Source transformations

4. Mesh Analysis

Mesh Analysis Recipe

1. Label each mesh (loop) in the circuit.

2. Label the currents flowing through each mesh .

3. Label the voltages across each component in the circuit.

4. For each mesh (loop):a) Use Kirchoff’s Voltage Law (KVL)

b) Use Ohm’s Law to substitute voltages for currents in KVL

5. Solve the set of equations for mesh currents.

6. Use mesh currents to calculate voltages for each component.

Mesh Analysis Example 1

Ω

ΩΩ

12 V 24 V6

93

+ -v1

+

-

v2

+ -v3

+

-

Ω

ΩΩ

12 V 24 V6

93

+ -v1

+

-

v2

+ -v3

KVL around Mesh 1 KVL around Mesh 2

+

-

Ohm’s LawVoltage drop across each resistor is the sum of the currents flowing through it.

𝑉 = 𝑖𝐴 + 𝑖𝐵 𝑅

𝑉 = 𝑖𝐴 + −𝑖𝐵 𝑅

𝑉 = 𝑖𝐴 −𝑖𝐵 𝑅

𝑖𝐵 is flowing from negative to positive so it must be negative!

Ω

ΩΩ

12 V 24 V6

93

+ -v1

+

-

v2

+ -v3

KVL around Mesh 1 KVL around Mesh 2

−12 + 3𝑖1 + 6(𝑖1 − 𝑖2) = 0 6 𝑖2 − 𝑖1 + 9𝑖2 + 24 = 0

+

-