Download I=1 A

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This lecture is brought to you by…
…Oreos.
Announcements
 Today is the last day to drop or change to hearer without
“WD” showing on transcript.*
 Please check the grades spreadsheet periodically for
accuracy. Let your recitation instructor know right away about
any discrepancies.
Spreadsheets will typically be updated on Sundays for the
rest of the semester (except for exam weeks).
*Not that I’m encouraging you to drop.
Today’s agenda:
Resistors in Series and Parallel.
You must be able to calculate currents and voltages in circuit components in series and in
parallel.
Kirchoff’s Rules.
You must be able to use Kirchoff’s Rules to calculate currents and voltages in circuit
components that are not simply in series or in parallel.
Resistances in Circuits
There are “two” ways to connect circuit elements.
Series:
A
B
Put your finger on the wire at A. If you can move along the
wires to B without ever having a choice of which wire to
follow, the circuit components are connected in series.
Truth in advertising: it is possible to have circuit elements that are connected
neither in series nor in parallel. See problem 24.73 in the 12th edition
of our text for an example with capacitors.
Parallel:
A
???
B
Put your finger on the wire at A. If in moving along the
wires to B you ever have a choice of which wire to follow,
the circuit components are connected in parallel.*
*Truth in advertising: actually, the circuit components are
not connected in series, and may be connected in parallel.
Are these resistors in series or parallel?
+ -
V
parallel
It matters where you put the source of emf.
Are these resistors in series or parallel?
+
V
series
It matters where you put the source of emf.
If resistors “see” the same potential difference, they are in
parallel. If resistors “see” the same current, they are in series.
V
+ -
+
I
V
V
parallel
series
It’s difficult to come up with a simple one- or two-sentence rule for series/parallel.
Here’s a circuit with three resistors and a battery:
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
Current flows…
…in the steady state, the same current flows through all
resistors…
…there is a potential difference (voltage drop) across each
resistor.
Applying conservation of energy allows us to calculate the
equivalent resistance of the series resistors.
I am including the derivation in these notes, for the benefit of
students who want to look at it.
In lecture, I will skip ahead past the derivation.
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
An electric charge q is given a potential energy qV by the
battery.
As it moves through the circuit, the charge loses potential
energy qV1 as it passes through R1, etc.
The charge ends up where it started, so the total energy lost
must equal the initial potential energy input:
qV = qV1 + qV2 + qV3 .
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
qV = qV1 + qV2 + qV3
V = V1 + V2 + V3
V = IR1 + IR2 + IR3
Now imagine replacing the three resistors by a single resistor,
having a resistance R such that it draws the same current as
the three resistors in series.
I
Req
V
+ -
I
As above:
From before:
Combining:
V
V = IReq
V = IR1 + IR2 + IR3
IReq = IR1 + IR2 + IR3
Req = R1 + R2 + R3
For resistors in series, the total resistance is the sum of the
separate resistances.
We can generalize this to any number of resistors:
R eq   R i
(resistors in series)
a consequence of
conservation of energy
i
R1
R2
R3
+ -
V
Note: for resistors in parallel, Req is always greater than any of the Ri.
Here’s another circuit with
three resistors and a battery.
I1
R1
V
I2
Current flows…
R2
V
…different currents flows through
different resistors…
R3
I3
V
…but the voltage drop across
each resistor is the same.
+ -
I
V
Applying conservation of charge allows us to calculate the
equivalent resistance of the parallel resistors.
I am including the derivation in these notes, for the benefit of
students who want to look at it.
In lecture, I will skip ahead past the derivation.
I1
In the steady state, the
current I “splits” into I1, I2,
and I3 at point A.
V
I2
A
I1, I2, and I3 “recombine” to
make a current I at point B.
B
R3
I3
I2 =
V
+ V
I
Because the voltage drop across
each resistor is V:
V
R1
R2
V
Therefore, the net current
flowing out of A and into B is I
= I1 + I2 + I3 .
I1 =
R1
V
R2
I3 =
V
R3
I
I
Now imagine replacing the
three resistors by a single
resistor, having a resistance R
such that it draws the same
current as the three resistors in
parallel.
Req
A
B
V
+ -
From above, I = I1 + I2 + I3, and
V
I1 =
R1
So that
V
I2 =
R2
V
V
V
V
= + + .
R eq R1 R 2 R 3
I
V
V
I3 =
.
R3
I
Dividing both sides by V gives
1
1
1
1
= + + .
R eq R1 R 2 R 3
We can generalize this to any number of resistors:
1
1

R eq
i Ri
(resistors in parallel)
a consequence of
conservation of charge
Note: for resistors in parallel, Req is always less than any of the Ri.
Summary:
Series
B
A
i
same I, V’s add
“just like” capacitors NOT
“just like” capacitors
Parallel
A
B
same V, I’s add
“just like” capacitors
R eq   R i
1
1

R eq
i Ri
“just like” capacitors NOT
Example: calculate the equivalent resistance of the resistor
“ladder” shown. All resistors have the same resistance R.
I’ll work this “conceptually.
A
B
Here’s the key to solving Physics problems: don’t bite off more
than you can chew. Bite off little bite-sized chunks.
A hot dog. Where do you take the first bite?
A
B
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
A deep philosophical question to consider. Do you always eat
your food from the outside in?
Example: For the circuit below, calculate the current drawn
from the battery and the current in the 6  resistor.
10 
8
6
8
1
3
9V
The next three slides show the strategy for solving this. I will
work the example at the blackboard in lecture. Skip to slide 37.
In a few minutes, 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
8
3
1
9V
Replace the parallel combination (green) by its equivalent.
Do you see any bite-sized chunks that are simple series or
parallel?
Any more “bite-sized chunks?” Remember that everything
inside the green box is equivalent to a single resistor.
10 
8
6
8
1
3
9V
Replace the series combination (blue box) by its equivalent.
We are left with an equivalent circuit of 3 resistors in series,
which is easy handle.
10 
8
6
8
1
3
9V
Next bite-sized chunk. Inside the blue box is “a” resistor.
Replace the parallel combination (orange) by its equivalent.
Let’s shrink the diagram a bit, and work this a step at a time.
R1=10 
R3 and R4 are in parallel.
R3=8 
R2=6 
R4=8 
R5=
3
1
1
1
1 1 2 1
=
+
= + = =
R 34 R 3 R 4 8 8 8 4
R 34 = 4 
R6=1  =9 V
R1=10 
R2 and R34 are in series.
R2=6 
R34=4 
R6=1  =9 V
R5=
3
R 234 = R 2 +R 34 = 6+4 =10 
Let’s shrink the diagram a bit, and work this a step at a time.
R1=10 
R1 and R234 are in parallel.
1
R234=10 
R5=
3
R1234
=
1
1
1
1
2 1
+
= + =
=
R1 R 234 10 10 10 5
R1234 = 5 
R6=1  =9 V
R1234, R5 and R6 are in series.
R1234=5 
R5=
3
R6=1  =9 V
R eq = R1234 +R 5 +R 6 = 5+3+1
R eq = 9 
Calculate the current drawn from the battery.
R eq = 9 
R1234=5 
R5=
3
I=1 A
R6=1  =9 V
V =I R

9
I=
= =1 A
R eq 9
Find the current in the 6  resistor.
R1=10 
R3=8 
R2=6 
I=1 A
R4=8 
R5=
3
There are many ways to
do the calculation. This is
just one.
R6=1  =9 V
R1=10 
R234=10 
I=1 A
R6=1  =9 V
R5=
3
V1 = V234 = V1234 (parallel).
Find the current in the 6  resistor.
I
V1234 = I R1234 = (1)(5) = 5 V
R1234=5 
R5=
3
I=1 A
V1 = V234 = 5 V
R6=1  =9 V
R1=10 
V234 = I234 R234
I234
R234=10 
I=1 A
R6=1  =9 V
R5=
3
I234 = V234 / R234 = 5/10
I234 = 0.5 A
Find the current in the 6  resistor.
R1=10 
R3=8 
I2=I234
R2=6 
I=1 A
R4=8 
R6=1  =9 V
I234 = I2 = I34 = 0.5 A
R5=
3
I2 = 0.5 A
Find the current in the 6  resistor.
R1=10 
R234=10 
I=1 A
R6=1  =9 V
A student who has taken
a circuits class will
probably say
R5=
3
R1 = R234
so I1 = I234 = I/2 = 0.5 A
If you want to do this on the exam, make sure you write down
your justification on the exam paper, and don’t make a
mistake! If you don’t show work and make a mistake, we can’t
give partial credit.
Answers without work shown generally receive no credit.
Example: two 100  light bulbs are connected (a) in series and
(b) in parallel to a 24 V battery. For which circuit will the bulbs
be brighter?
1.
2.
parallel (left)
series (right)
Example: two 100  light bulbs are connected (a) in series and
(b) in parallel to a 24 V battery. What is the current through
each bulb? For which circuit will the bulbs be brighter?
(a) Series combination.
Req = R1 + R2
R1
100
R2
100
+ -
V = I Req
I
V = 24 V
V = I (R1 + R2)
I = V / (R1 + R2) = 24 V / (100  + 100 ) = 0.12 A
Example: two 100  light bulbs are connected (a) in series and
(b) in parallel to a 24 V battery. What is the current through
each bulb? For which circuit will the bulbs be brighter?
I1
(b) Parallel combination.
1
1
1
1
1
2
=
+
=
+
=
R eq
R1
R2
100 100 100
R1
V
I2
R2
V
R eq = 50 
V = I R eq  I = V
I =
+ -
R eq
24
= 0.48 A
50
I1 = I2 = I
2
= 0.24 A
I
V = 24 V
(because R1 = R2)
I
Example: two 100  light bulbs are connected (a) in series and
(b) in parallel to a 24 V battery. What is the current through
each bulb? For which circuit will the bulbs be brighter?
To answer the question, we must calculate the power dissipated
in the bulbs for each circuit. The more power “consumed,” the
brighter the bulb.
In other words, we use power as a proxy for brightness.
Example: two 100  light bulbs are connected (a) in series and
(b) in parallel to a 24 V battery. What is the current through
each bulb? For which circuit will the bulbs be brighter?
(a) Series combination.
We know the resistance and
current through each bulb, so for
each bulb:
P=
I2R
P = (0.12 A)2 (100 )
P = 1.44 W
R1
100
R2
100
+ -
I
V = 24 V
Example: two 100  light bulbs are connected (a) in series and
(b) in parallel to a 24 V battery. What is the current through
each bulb? For which circuit will the bulbs be brighter?
(b) Parallel combination.
We know the potential difference
across each bulb, so for each
bulb:
P = V2 / R
P = (24 V)2 / ( 100 )
I1
R1
V
I2
I
R2
V
I
+ -
V = 24 V
P = 5.76 W
We also know each current, so we could have used P = I2R.
Example: two 100  light bulbs are connected (a) in series and
(b) in parallel to a 24 V battery. What is the current through
each bulb? For which circuit will the bulbs be brighter?
I1
R1
100
R2
100
+ -
I
V = 24 V
R1
V
I2
I
R2
V
+ -
V = 24 V
Compare:
Pseries = 1.44 W
Pparallel = 5.76 W
The bulbs in parallel are brighter.
I
This is what you see if you connect 40 W bulbs directly to a 120
V outlet. (DO NOT TRY AT HOME.)
Off
On
Homework Hints
(usefulness depends on what homework is assigned this semester)
Power = Energy Transformed / Time
Energy Transformed = Power * Time
Do you remember this Phys. 1135 equation?
Q (energy) = m c T
A Toy to Play With
http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc
Today’s agenda:
Resistors in Series and Parallel.
You must be able to calculate currents and voltages in circuit components in series and in
parallel.
Kirchoff’s Rules.
You must be able to use Kirchoff’s Rules to calculate currents and voltages in circuit
components that are not simply in series or in parallel.
Kirchhoff’s Rules
No, it is pronounced “KEERKOFF’s” rules. The ch sounds like
“k,” not like “ch.” (depending on regional accents)
Analyze this circuit for me, please. Find the currents I1, I2, and
I3.
h
30 
I1
I3
40 
a
1
2 = 45 V
c
b
d
20 
I2
1 = 85 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? We haven’t
discussed how to analyze that combination.
h
30 
I1
I3
40 
a
1
2 = 45 V
c
b
series1
I2
1 = 85 V 1 
g
f
series2
e
d
20 
A new technique is needed to analyze this, and far more
complex circuits.
Kirchhoff’s Rules
Kirchhoff’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 Kirchhoff’s First Rule.*
Kirchhoff’s Loop Rule: the sum of the changes of potential
around any closed path of a circuit must be zero. Also called
Kirchhoff’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 ).
Kirchhoff’s Rules
Starting Equations
I =0
V =0
at any junction
around any closed loop
You already used
this in your
homework!
simple… but there are details to worry about…
If this were Physics 1135, you would have a Kirchhoff’s Rules litany.
You have my permission to print this page and use it for tomorrow’s boardwork.
Brief litany for Kirchhoff’s Rules 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 (OK to guess direction).
4. Apply Kirchhoff’s Junction Rule at each junction.
Current in is +.
5. Apply Kirchhoff’s Loop Rule for as many loops as
necessary. Follow each loop in one direction only.
+
5a. Resistor: I
5b. Battery:
V is +
V is loop
loop
6. Solve.
h
30 
I1
I3
40 
a
1
2 = 45 V
c
b
dd
20 
I2
1 = 85 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 = 85 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.
h
30 
I1
I3
40 
a
1
2 = 45 V
c
b
d
20 
I2
1 = 85 V
g
1
e
f
The “green” loop (a-h-d-c-b-a):
(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0
I
V is loop
+-
V is +
loop
h
30 
I1
I3
40 
a
1
2 = 45 V
c
b
d
20 
I2
1 = 85 V
g
1
e
f
The “blue” loop (a-b-c-d-e-f-g):
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
I
V is loop
+-
V is +
loop
After combining terms and simplifying, we now have three
equations, three unknowns; the rest is “just algebra.”
Junction a:
I3 – I1 – I 2 = 0
--eq. 1
The “green” loop
- 30 I1 + 45 - 41 I3 = 0
--eq. 2
The “blue” loop
41 I3 -130 + 21 I2 = 0
--eq. 3
skip the algebra!
Make sure to use voltages in V and resistances in . Then currents will be in A.
You can see the solution in part 7 of today’s lecture notes (which will not be
presented in class). It’s a pain. Generally, with the homework problems, you can
easily solve directly for one variable, leaving you with two equations and two
unknowns.
Don’t forget…
eat your oreos!