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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
editionof 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?
It matters where you put the source of emf.
+ -
V
parallel
Are these resistors in series or parallel?
It matters where you put the source of emf.
+- V
series
If resistors “see” the same potential difference, they are in parallel. If resistors “see” the same current, they are in series.
It’s difficult to come up with a simple one- or two-sentence rule for series/parallel.
+- V
series
+ -
V
parallel
I
V
Here’s a circuit with three resistors and a battery:
R3R2R1
+ -
VI
Current flows…
…in the steady state, the same current flows through all resistors…
III
…there is a potential difference (voltage drop) across each resistor.
V1 V3V2
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.
An electric charge q is given a potential energy qV by the battery.
R3R2R1
+ -
VI
III
V1 V3V2
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 .
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.
R3R2R1
+ -
VI
III
V1 V3V2
qV = qV1 + qV2 + qV3
Req
+ -
VI
V
I
As above: V = IReq
From before: V = IR1 + IR2 + IR3
Combining: 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:
(resistors in series)a consequence of conservation of energy
R3R2R1
+ -
V
eq ii
R R
Note: for resistors in parallel, Req is always greater than any of the Ri.
V
V
V
R3
R2
R1
+ -
VI
Current flows…
…different currents flows through different resistors…
…but the voltage drop across each resistor is the same.
I3
I1
I2
Here’s another circuit with three resistors and a battery.
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.
V
V
V
R3
R2
R1
+ -
VI
I3
I1
I2
A B
In the steady state, the current I “splits” into I1, I2, and I3 at point A.
I
I1, I2, and I3 “recombine” to make a current I at point B.
Therefore, the net current flowing out of A and into B is I = I1 + I2 + I3 .
1 2 31 2 3
V V VI = I = I =
R R R
Because the voltage drop across each resistor is V:
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.
V
Req
+ -
VI
I
A B
IFrom above, I = I1 + I2 + I3, and
1 2 31 2 3
V V VI = I = I = .
R R R
So thateq 1 2 3
V V V V = + + .
R R R R
Dividing both sides by V gives
eq 1 2 3
1 1 1 1 = + + .
R R R R
We can generalize this to any number of resistors:
(resistors in parallel)a consequence of conservation of chargeieq i
1 1
R R
Note: for resistors in parallel, Req is always less than any of the Ri.
Summary:
Series A Beq i
i
R Rsame I, V’s add
Parallel A B
same V, I’s add
ieq i
1 1
R R
“just like” capacitors
“just like” capacitors
“just like” capacitors NOT
“just like” capacitors NOT
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.
Example: calculate the equivalent resistance of the resistor “ladder” shown. All resistors have the same resistance R.
A
B
A hot dog. Where do you take the first bite?
A
B
Series
Not a “law” of physics, but sometimes helps with circuits: look for “bite-sized” chunks sticking out at one end.
A
B
Parallel
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.
Any more bite-sized chunks?
A
B
Series
A
B
Parallel
A
B
Series
A
B
All done!
A deep philosophical question to consider. Do you always eat your food from the outside in?
3 8
8
10
6
1 9 V
The next three slides show the strategy for solving this. I will work the example at the blackboard in lecture. Skip to slide 37.
Example: For the circuit below, calculate the current drawn from the battery and the current in the 6 resistor.
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.”
Replace the parallel combination (green) by its equivalent.
3 8
8
10
6
1 9 V
Do you see any bite-sized chunks that are simple series or parallel?
8 3
8
10
6
1 9 V
Replace the series combination (blue box) by its equivalent.
Any more “bite-sized chunks?” Remember that everything inside the green box is equivalent to a single resistor.
8 3
8
10
6
1 9 V
We are left with an equivalent circuit of 3 resistors in series, which is easy handle.
Next bite-sized chunk. Inside the blue box is “a” resistor. Replace the parallel combination (orange) by its equivalent.
R5=3
R1=10
R2=6
R6=1 =9 V
R4=8
R3=8
Let’s shrink the diagram a bit, and work this a step at a time.
R3 and R4 are in parallel.
34 3 4
1 1 1 1 1 2 1 = + = + = =
R R R 8 8 8 4
34R = 4
R1=10
R2=6
R6=1 =9 V
R34=4
R2 and R34 are in series.
234 2 34R = R +R =6+4=10 R5=3
Let’s shrink the diagram a bit, and work this a step at a time. R1=10
R6=1 =9 V
R234=10 R5=3
R1 and R234 are in parallel.
1234 1 234
1 1 1 1 1 2 1 = + = + = =
R R R 10 10 10 5
1234R = 5
R6=1 =9 V
R1234=5
R5=3
R1234, R5 and R6 are in series.
6eq 1234 5R = R +R +R =5+3+1
eqR =9
Calculate the current drawn from the battery.
R6=1 =9 V
R1234=5
R5=3
eqR =9
V =I R
eq
9I = = =1 A
R 9I=1 A
R5=3
R1=10
R2=6
R6=1 =9 V
R4=8
R3=8
Find the current in the 6 resistor.
There are many ways to do the calculation. This is just one.
I=1 A
R1=10
R6=1 =9 V
R234=10 R5=3
V1 = V234 = V1234 (parallel).
I=1 A
Find the current in the 6 resistor.
V1234 = I R1234 = (1)(5) = 5 V
R1=10
R6=1 =9 V
R234=10 R5=3
V234 = I234 R234
R6=1 =9 V
R1234=5
R5=3
I=1 A
I=1 A
I234 = V234 / R234 = 5/10
I234 = 0.5 A
I
I234
V1 = V234 = 5 V
Find the current in the 6 resistor.
I234 = I2 = I34 = 0.5 A
R5=3
R1=10
R2=6
R6=1 =9 V
R4=8
R3=8
I=1 A
I2 = 0.5 A
I2=I234
Find the current in the 6 resistor.
A student who has taken a circuits class will probably say
R1=10
R6=1 =9 V
R234=10 R5=3
I=1 A
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. parallel (left)
2. series (right)
(a) Series combination. R1
100+ -
I V = 24 V
Req = R1 + R2
V = I Req
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?
R2
100
(b) Parallel combination. V
V
R2
R1
+ -
V = 24 VI
I2
I1
I
eq 1 2
1 1 1 1 1 2 = + = + =
R R R 100 100 100
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?
eqR = 50
eqeq
VV = I R I = R
24I = = 0.48 A
50
1 2II = I = = 0.24 A2 (because R1 = R2)
To answer the question, we must calculate the power dissipated in the bulbs for each circuit. The more power “consumed,” the brighter the bulb.
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?
In other words, we use power as a proxy for brightness.
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
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. R1
100+ -
I V = 24 V
R2
100
We know the potential difference across each bulb, so for each bulb:
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. V
V
R2
R1
+ -
V = 24 V
I
I2
I1
IP = V2 / R
P = (24 V)2 / ( 100 )
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?
V
V
R2
R1
+ -
V = 24 V
I
I2
I1
I
Pparallel = 5.76 W
The bulbs in parallel are brighter.
Compare: Pseries = 1.44 W
R1
100+ -
I V = 24 V
R2
100
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 Withhttp://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.
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.
1 1 = 85 V
1 2 = 45 V
20
30
40 a
b cd
eg f
h
I3
I2
I1
Kirchhoff’s Rules
1 1 = 85 V
1 2 = 45 V
20
30
40 a
b cd
eg f
h
I3
I2
I1
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.
series1
series2
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
at any junction I =0
around any closed loopV=0
simple… but there are details to worry about…
You already used this in
your homework!
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 +.
Brief litany for Kirchhoff’s Rules Problems
5. Apply Kirchhoff’s Loop Rule for as many loops as necessary. Follow each loop in one direction only.
5a. Resistor: +-I
loop
V is -5b. Battery:
loop
V is +
6. Solve.
You have my permission to print this page and use it for tomorrow’s boardwork.
If this were Physics 1135, you would have a Kirchhoff’s Rules litany.
1 1 = 85 V
1 2 = 45 V
20
40 a
b cd
eg f
h
I3
I2
I1
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 – I2 = 0 --eq. 1
Junction d: -I3 + I1 + I2 = 0
Junction d gave no new information, so we still need two more equations.
30
da
1 1 = 85 V
1 2 = 45 V
20
40 a
b cd
eg f
h
I3
I2
I1
There are three loops.
Loop 1. Loop 2. Loop 3.
Any two loops will produce independent equations. Using the third loop will provide no new information.
30
1 1 = 85 V
1 2 = 45 V
20
40 a
b cd
eg f
h
I3
I2
I130
The “green” loop (a-h-d-c-b-a):
(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0
I
loop
V is -+-
loop
V is +
1 1 = 85 V
1 2 = 45 V
20
40 a
b cd
eg f
h
I3
I2
I130
The “blue” loop (a-b-c-d-e-f-g):
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
I
loop
V is -+-
loop
V is +
The “green” loop - 30 I1 + 45 - 41 I3 = 0 --eq. 2
The “blue” loop 41 I3 -130 + 21 I2 = 0 --eq. 3
After combining terms and simplifying, we now have three equations, three unknowns; the rest is “just 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.
skip the algebra!
Junction a: I3 – I1 – I2 = 0 --eq. 1
Don’t forget…
eat your oreos!
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