PHY 202 (Blum) 1

Kirchhoff’s Rules

When series and parallel combinations aren’t enough

2PHY 202 (Blum)

Some circuits have resistors which are neither in series nor parallel

They can still be analyzed, but one uses Kirchhoff’s rules.

3PHY 202 (Blum)

Not in series

The 1-k resistor is not in series with the 2.2-k since the some of the current that went through the 1-k might go through the 3-k instead of the 2.2-k resistor.

4PHY 202 (Blum)

Not in parallel

The 1-k resistor is not in parallel with the 1.5-k since their bottoms are not connected simply by wire, instead that 3-k lies in between.

5PHY 202 (Blum)

Kirchhoff’s Node Rule

A node is a point at which wires meet. “What goes in, must come out.” Recall currents have directions, some currents will point

into the node, some away from it. The sum of the current(s) coming into a node must equal

the sum of the current(s) leaving that node.

I1 + I2 = I3 I1 I2

I3The node rule is about currents!

6PHY 202 (Blum)

Kirchhoff’s Loop Rule 1 “If you go around in a circle, you get back

to where you started.” If you trace through a circuit keeping track

of the voltage level, it must return to its original value when you complete the circuit

Sum of voltage gains = Sum of voltage losses

7PHY 202 (Blum)

Batteries (Gain or Loss)

Whether a battery is a gain or a loss depends on the direction in which you are tracing through the circuit

Gain Loss

Loo

p di

rect

ion

Loo

p di

rect

ion

8PHY 202 (Blum)

Resistors (Gain or Loss)

Whether a resistor is a gain or a loss depends on whether the trace direction and the current direction coincide or not.

I I

Loss Gain

Loo

p di

rect

ion

Loo

p di

rect

ion

Cur

rent

dir

ectio

n

Cur

rent

dir

ectio

n

PHY 202 (Blum) 9

Branch version

10PHY 202 (Blum)

Neither Series Nor Parallel

I1

I2.2

I1.5

I1.7

I3

Assign current variables to each branch. Draw loops such that each current element is included in at least one loop.

11PHY 202 (Blum)

Apply Current (Node) Rule

I1

I1-I3

I1.5

I1.5+I3

I3

*Node rule applied.

* *

12PHY 202 (Blum)

Three Loops

Voltage Gains = Voltage Losses 5 = 1 • I1 + 2.2 • (I1 – I3)

1 • I1 + 3 • I3 = 1.5 • I1.5

2.2 • (I1 – I3) = 3 • I3 + 1.7 • (I1.5 + I3)

Units: Voltages are in V, currents in mA, resistances in k

13PHY 202 (Blum)

5 = 1 • I1 + 2.2 • (I1 – I3)

I1

I1-I3

I1.5

I1.5+I3

I3

14PHY 202 (Blum)

1 • I1 + 3 • I3 = 1.5 • I1.5

I1

I1-I3

I1.5

I1.5+I3

I3

15PHY 202 (Blum)

2.2 • (I1 – I3) = 3 • I3 + 1.7 • (I1.5 + I3)

I1

I1-I3

I1.5

I1.5+I3

I3

16PHY 202 (Blum)

Simplified Equations

5 = 3.2 • I1 - 2.2 • I3

I1 = 1.5 • I1.5 - 3 • I3

0 = -2.2 • I1 + 1.7 • I1.5 + 6.9 • I3

Substitute middle equation into others

5 = 3.2 • (1.5 • I1.5 - 3 • I3) - 2.2 • I3

0 = -2.2 • (1.5 • I1.5 - 3 • I3) + 1.7 • I1.5 + 6.9 • I3

Multiply out parentheses and combine like terms.

17PHY 202 (Blum)

Solving for I3

5 = 4.8 • I1.5 - 11.8 • I3

0 = - 1.6 I1.5 + 13.5 • I3

Solve the second equation for I1.5 and substitute that result into the first

5 = 4.8 • (8.4375 I3 ) - 11.8 • I3

5 = 28.7 • I3

I3 0.174 mA

18PHY 202 (Blum)

Comparison with Simulation

19PHY 202 (Blum)

Other currents

Return to substitution results to find other currents.

I1.5 = 8.4375 I3 = 1.468 mA

I1 = 1.5 • I1.5 - 3 • I3

I1 = 1.5 • (1.468) - 3 • (0.174)

I1 = 1.68 mA

PHY 202 (Blum) 20

Loop version

21PHY 202 (Blum)

Neither Series Nor Parallel

Draw loops such that each current element is included in at least one loop. Assign current variables to each loop. Current direction and lop direction are the same.

JA

JB

JC

22PHY 202 (Blum)

Loop equations

5 = 1 (JA - JB) + 2.2 (JA - JC)

0 = 1 (JB - JA) + 1.5 JB + 3 (JB - JC)

0 = 2.2 (JC - JA) + 3 (JC - JB) + 1.7 JC

“Distribute” the parentheses

5 = 3.2 JA – 1 JB - 2.2 JC

0 = -1 JA + 5.5 JB – 3 JC

0 = -2.2JA – 3 JB + 6.9 JC

23PHY 202 (Blum)

Algebra

JC = (2.2/6.9)JA + (3/6.9)JB

JC = 0.3188 JA + 0.4348 JB

5 = 3.2 JA – 1 JB - 2.2 (0.3188 JA + 0.4348 JB) 0 = -1 JA + 5.5 JB – 3 (0.3188 JA + 0.4348 JB)

5 = 2.4986 JA – 1.9566 JB

0 = -1.9564 JA + 4.1956 JB

24PHY 202 (Blum)

More algebra

JB = (1.9564/4.1956) JA

JB = 0.4663 JA

5 = 2.4986 JA – 1.9566 (0.4663 JA)

5 = 1.5862 JA

JA = 3.1522 mA

25PHY 202 (Blum)

Other loop currents

JB = 0.4663 JA = 0.4663 (3.1522 mA)

JB = 1.4699 mA

JC = 0.3188 JA + 0.4348 JB

JC = 0.3188 (3.1522) + 0.4348 (1.4699)

JC = 1.644 mA

26PHY 202 (Blum)

Branch Variables

I1

I2.2

I1.5

I1.7

I3

Assign current variables to each branch. Draw loops such that each current element is included in at least one loop.

27PHY 202 (Blum)

Loop Variables

Draw loops such that each current element is included in at least one loop. Assign current variables to each loop. Current direction and lop direction are the same.

JA

JB

JC

28PHY 202 (Blum)

Branch Currents from Loop currents

I1 = JA – JB = 3.1522 – 1.4699 = 1.6823 mA

I1.5 = JB = 1.4699 mA

PHY 202 (Blum) 29

Matrix equation

30PHY 202 (Blum)

Loop equations as matrix equation

5 = 3.2 JA – 1 JB - 2.2 JC

0 = -1 JA + 5.5 JB – 3 JC

0 = -2.2JA – 3 JB + 6.9 JC

0

0

5

9.632.2

35.51

2.212.3

C

B

A

J

J

J

31PHY 202 (Blum)

Enter matrix in Excel, highlight a region the same size as the matrix.

32PHY 202 (Blum)

In the formula bar, enter =MINVERSE(range) where range is the set of cells corresponding to the matrix (e.g. B1:D3). Then hit Crtl+Shift+Enter

33PHY 202 (Blum)

Result of matrix inversion

34PHY 202 (Blum)

Prepare the “voltage vector”, then highlight a range the same size as the vector and enter =MMULT(range1,range2) where range1 is the inverse matrix and range2 is the voltage vector. Then Ctrl-Shift-Enter.

Voltage vector

35PHY 202 (Blum)

Results of Matrix Multiplication