Transcript lec26

The “bite-sized chunk” technique is a good one for any
problem.
Break a complex system into simple components and do
them one at a time.
Your problem-solving prowess will be tested most severely
when you encounter a problem for which this technique
won’t work.
We are about to encounter such a problem, in an electrical
circuit.
Due to the lack of time, I am going to once again set a bad
example and just “cover” the technique.
Kirchoff’s Rules
No, it is pronounced “KIRKOFF’s” rules. The ch sounds
like “k,” not like “ch.”
Analyze this circuit for me, please. Find the currents I1, I2,
and I3. Be flexible: I could give I1 instead of 2, and ask for I2, I3, & 2.
h
30 
I1
40 
I3
a
1
2 = 45 V
c
b
d
20 
I2
1 = 80 V
g
f
1
e
I see two sets of resistors in series. This.
And this.
You know how to analyze those.
Further analysis is difficult. For example, series1 seems to
be in parallel with the 30  resistor, but what about 2? You
don’t know how to analyze that.
h
30 
I1
40 
I3
a
1
2 = 45 V
c
b
series1
I2
1 = 80 V
g
f
1
d
20 
series2
e
A new technique is needed to analyze this, and far more
complex circuits.
Kirchoff’s Rules
Kirchoff’s Junction Rule: at any junction point, the sum of
all currents entering the junction must equal the sum of all
currents leaving the junction. Also called Kirchoff’s First
Rule.*
Kirchoff’s Loop Rule: the sum of the changes of potential
around any closed path of a circuit must be zero. Also
called Kirchoff’s Second Rule.**
*This is just conservation of charge: charge in = charge out.
**This is just conservation of energy: a charge ending up where it
started out neither gains nor loses energy (Ei = Ef ).
Solving Problems with Kirchoff’s Rules
Just as we had a litany for force problems in our Mechanics
section, we have a litany for circuit problems.
Litany for Circuit Problems
1. Draw the circuit (if not already done).
2. Label + and – for each battery (the short side is -).
3. Label the current in each branch of the circuit with a
symbol and an arrow. You may choose whichever
direction you wish for the arrow.
4. Apply Kirchoff’s Junction Rule at each junction. The
direction of the current arrows tell you whether current
is flowing in (+) or out (-).
Step 4 will probably give you fewer equations than variables. Proceed
to step 5 go get additional equations.
5. Apply Kirchoff’s Loop Rule for as many loops as
necessary to get enough equations to solve for your
unknowns. Follow each loop in one direction only—
your choice.
5a. For a resistor, the sign of the potential
difference is negative if your chosen loop
direction is the same as the chosen current
direction through that resistor; positive if
opposite.
5b. For a battery, the sign of the potential
difference is positive if your chosen loop
direction moves from the negative terminal
towards the positive; negative if opposite.
6. Collect equations, solve, and check results.
We need a shortened version of the litany for quick
reference.
Brief litany for Circuit Problems
1. Draw the circuit.
2. Label + and – for each battery.
3. Label the current in each branch of the circuit with a
symbol and an arrow.
4. Apply Kirchoff’s Junction Rule at each junction.
Current in is +.
5. Apply Kirchoff’s Loop Rule for as many loops as
necessary. Follow each loop in one direction only.
5a. Resistor: I
V is loop
6. Solve.
5b. Battery:
+-
V is +
loop
h
30 
I1
40 
I3
a
1
2 = 45 V
c
b
d
20 
I2
1 = 80 V
g
f
1
e
Back to our circuit: we have 3 unknowns (I1, I2, and I3), so
we will need 3 equations. We begin with the junctions.
Junction a:
I3 – I 1 – I2 = 0
Junction d:
-I3 + I1 + I2 = 0
--eq. 1
Junction d gave no new information, so we still need two more equations.
h
30 
I1
40 
I3
a
1
2 = 45 V
c
b
d
20 
I2
1 = 80 V
g
There are three loops.
f
1
e
Loop 1. Loop 2. Loop 3.
Any two loops will produce independent equations.
Using the third loop will provide no new information.
Reminders:
I
+-
V is loop
V is +
loop
The “green” loop:
(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0
- 30 I1 + 45 - 41 I3 = 0
--eq. 2
The “blue” loop:
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
41 I3 -130 + 21 I2 = 0
--eq. 3
Three equations, three unknowns; the rest is “algebra.”
Make sure to use voltages in V and resistances in . Then currents will be A.
Collect our three equations:
I3 – I1 – I2 = 0
- 30 I1 + 45 - 41 I3 = 0
41 I3 -130 + 21 I2 = 0
Rearrange to get variables in “right” order:
– I 1 – I 2 + I3 = 0
- 30 I1 - 41 I3 + 45 = 0
21 I2 + 41 I3 -130 = 0
Use the middle equation to eliminate I1:
I1 = (41 I3 – 45)/(-30)
There are many valid sets of steps to solving a system of equations.
Any that works is acceptable.
Two equations left to solve:
– (41 I3 – 45)/(-30) – I2 + I3 = 0
21 I2 + 41 I3 -130 = 0
Might as well work out the numbers:
1.37 I3 – 1.5 – I2 + I3 = 0
21 I2 + 41 I3 -130 = 0
– I2 + 2.37 I3 – 1.5 = 0
21 I2 + 41 I3 -130 = 0
Multiply the top equation by 21:
– 21 I2 + 49.8 I3 – 31.5 = 0
21 I2 + 41 I3 -130 = 0
Add the two equations to eliminate I2:
– 21 I2 + 49.8 I3 – 31.5 = 0
+ ( 21 I2 + 41 I3 -130 = 0 )
90.8 I3 – 161.5 = 0
Solve for I3:
I3 = 161.5 / 90.8
I3 = 1.78
Go back to the “middle equation” two slides ago for I1:
I1 = (41 I3 – 45)/(-30)
I1 = - 1.37 I3 + 1.5
I1 = - (1.37) (1.78) + 1.5
I1 = - 0.94
Go back two slides to get an equation that gives I2:
– I2 + 2.37 I3 – 1.5 = 0
I2 = 2.37 I3 – 1.5
I2 = (2.37) (1.78) – 1.5
I2 = 2.72
Summarize answers so your lazy prof doesn’t have to go
searching for them and get irritated (don’t forget to show
units in your answer):
I1 = - 0.94 A
I2 = 2.72 A
I3 = 1.78 A
These don’t look “quite like” the text answer. They rounded
to 2 digits. Maybe that’s why. Better check anyway.
I3 – I1 – I2 = 0
- 30 I1 + 45 - 41 I3 = 0
41 I3 -130 + 21 I2 = 0
I1 = - 0.94 A
I2 = 2.72 A
I3 = 1.78 A
1.78 – (-0.94) – 2.72 = 0

- 30 (-0.94) + 45 - 41 (1.78) = 0.22
?
41 (1.78) -130 + 21 (2.72) = 0.10
?
Are the last two indication of a mistake or just roundoff error?
Recalculating while retaining 2 more digits gives I1=0.933, I2=2.714,
I3=1.7806, and the last two results are 0.01 or less  roundoff was the
culprit.
Even a relatively simple circuit can easily produce a 5x5 or
much greater system of equations to be solved.
See here for an example solved homework problem. Good
luck solving a problem like this on an exam!
Assorted Topics
…I probably won’t have time to get to.
You can connect batteries in series and parallel, just like
resistors. Voltages add for batteries in series. Not so for
batteries in parallel—use Kirchoff’s rules to analyze.
You can connect a battery in series the “wrong way” with
other batteries—useful for charging a rechargeable.
Go to www.howstuffworks.com to see how batteries work.
They even expose the secret of the 9 volt battery!
Click on the picture above only if you are mature enough
to handle this graphic exposé.
Go to www.howstuffworks.com to see how batteries work.
They even expose the secret of the 9 volt battery!
Shocking!
Six 1.5 V batteries in series!
You can also connect capacitors in series and in parallel.
C1
Ceq = Ci
C2
a
(capacitors in parallel)
C2
+ V
C1
1
1
=
(capacitors in series)
Ceq
Ci
i
C2
+ V
Look familiar?
C3