Transcript Example 19-6

Upcoming Schedule
Oct. 8
19.2-19.4
Oct. 10
boardwork
Quiz 4
Oct. 13
19.5-19.9
Oct. 15
boardwork
Oct. 17
boardwork
Quiz 5
Oct. 20
review
Oct. 22
Exam 2
Ch. 18, 19
Oct. 24
20.1-20.2
Oct. 6
19.1
“As far as the laws of mathematics refer to reality, they are not certain, and
as far as they are certain, they do not refer to reality.”—A. Einstein
Examples 19-3 and 19-4
How much current flows from the battery in the circuit shown?
What is the current through the 500  resistor?
500 
400 
a
I1
c
b
700 
I
I2
+ -
12 V
I
What is your stragegy? Step 1—replace the 500 and 700 
parallel combination by a single equivalent resistor.
500 
400 
a
I1
I=?
c
b
700 
I
I2
I
I1 = ?
+ -
12 V
Woe is me, what to do? Always think: bite-sized chunks!
Step 2—replace the 400 and Req1 series combination by a single
equivalent resistor Req, net.
Req1
400 
a
I=?
c
b
I
I
I1 = ?
+ -
12 V
Woe is me, what to do? Find another bite-sized chunk!
Step 3—Solve for the current I.
Req1, net
c
a
I
I
+ -
12 V
This isn’t so complicated!
Step 4—To get I1, Calculate Vbc.
Use Vtotal = Vab + Vbc.
Vbc
500 
Vab
Knowing I,
Calculate I1.
Woe is me!
Stuck
again!
400 
a
I1
c
b
700 
I
I2
I
+ -
12 V
You know Vtotal= V and I so you can get Vab and then Vbc.
The voltage drop across both the 500 and 700  resistors is the
same, and equal to Vbc. Use V = IR to get I1 across the 500 
resistor.
500 
400 
a
I1
c
b
700 
I
I2
+ -
12 V
I
I’ll post some sample solutions, worked out in detail, for similar
problems. The previous slides were intended to illustrate
concepts and techniques only.
The next slides on the “resistor ladder” are also conceptual
only.
Don’t worry, there will be plenty of practice with “real”
problems.
Example 19-6
Resistor “ladder.” Estimate the equivalent resistance of the
resistor ladder shown. All resistors have the same resistance R.
A
B
Remember this (seriously): don’t bite off more than you can
chew. Bite off little bite-sized chunks.
Not a “law” of physics, but sometimes helps with circuits: look
for “bite-sized” chunks sticking out at one end.
Series
A
B
The new color indicates the value of the resistance is not R. In
a real problem, you would calculate the “new color” resistor’s
resistance.
Parallel
A
B
Any more bite-sized chunks?
Series
A
B
Parallel
A
B
Series
A
B
All done!
A
B
19.2 EMF and Terminal Voltage
We have been making calculations with voltages from batteries
without asking detailed questions about the batteries. Now it’s
time to ask those questions.
We introduce a new term – emf – in this section.
Any device which transforms a form of energy into electric
energy is called a “source of emf.”
“emf” is an abbreviation for “electromotive force,” but emf does
not really refer to force!
The emf of a source is the voltage it produces when no current
is flowing.
The voltage you measure across the terminals of a battery (or
any source of emf) is less than the emf because of internal
resistance.
Here’s a battery with an emf. Also, all batteries have an
emf is the potential difference between + and “internal resistance:”
a
+ -
b
The “battery” is everything
inside the green box.
Hook up a voltmeter to measure the emf:
emf
a
+ -
b
The “battery” is everything
inside the green box.
Getting ready to connect the
voltmeter (it’s not hooked up
yet).
Measuring the emf???
a
emf
+ -
I
b
The “battery” is everything
inside the green box.
As soon as you connect the
voltmeter, current flows.
You can’t measure voltage without some (however
small) current flowing, so you can’t measure emf
directly.
You can only measure Vab.
We model a battery as producing an emf, , and having an
internal resistance r:
r
a
+ -

r
b
The “battery” is everything
inside the green box.

Vab
The terminal voltage, Vab, is the voltage you measure with
current flowing. When a current I flows through the battery,
Vab is related to the emf, , by
Vab = ε ± I r .
Why the  sign? If the battery is delivering current, the V it
delivers is less than the emf, so the – sign is necessary.
If the battery is being charged, you have to “force” the current
through the battery, and the V to “force” the current through is
greater than the emf, so the + sign is necessary.
This will become clear as you work (and understand) problems.
Operationally, you simply include an extra resistor to represent
the battery resistance, and label the battery voltage as an emf
instead of V (units are still volts).
Example 19-7
For the circuit below, calculate the current drawn from the
battery, the terminal voltage of the battery, and the current in
the 6  resistor.
10 
8
6
4
5
0.5   = 9 V
The following is a “conceptual” solution. Please go back and
put in the numbers for yourself.
In the next section, we will learn a general technique for solving
circuit problems. For now, we break the circuit into
manageable bits. “Bite-sized chunks.”
10 
8
6
4
5
0.5   = 9 V
Replace the parallel combination by its equivalent.
Do you see any bite-sized chunks that are simple series or
parallel?
Any more “bite-sized chunks?” Pretend that everything inside
the green box is a single resistor.
10 
8
6
4
0.5   = 9 V
Replace the series combination by its equivalent.
5
You are left with an equivalent circuit of 3 resistors in series,
which you can handle.
10 
8
6
4
5
0.5   = 9 V
Next bite-sized chunk. Inside the blue box is “a” resistor.
Replace the parallel combination by its equivalent.
19.3 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. (I could give I1 instead of 2, and ask for I2, I3, & 2. Lots of combinations.)
h
30 
I1
I3
40 
a
1
2 = 45 V
c
b
d
20 
I2
1 = 80 V 1 
g
f
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 combination.
h
30 
I1
I3
40 
a
1
2 = 45 V
c
b
series1
I2
1 = 80 V 1 
g
f
series2
e
d
20 
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 ).
19.4 Solving Problems with Kirchoff’s Rules
Just as we had a litany for force problems in our Mechanics
semester, we have a litany for circuit problems.
Litany for Circuit Problems
1. Draw the circuit.
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
I3
40 
a
1
2 = 45 V
c
b
d
20 
I2
1 = 80 V
g
1
f
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 – I1 – I 2 = 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
I3
40 
a
1
2 = 45 V
c
b
d
20 
I2
1 = 80 V
g
There are three loops.
1
e
f
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 (a-h-d-c-b-a):
(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0
- 30 I1 + 45 - 41 I3 = 0
--eq. 2
The “blue” loop (a-b-c-d-e-f-g):
(+ 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 are in V and resistances in . Then currents will be in 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:
– I1 – I2 + 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” Giancoli’s. He rounded to 2
digits. Maybe that’s why. Still, we’d better check our results.
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.