11/3/2015
Electric circuits
Summary
Tuesday’s lecture:
• Discussed role of wires, battery and bulb in a simple circuit
- What are the electrons doing in each part of the circuit
• Introduced resistance (and electron man)
Today’s lecture:
• Ohm’s Law
• Power formula
Lecture 22 :
Electric circuits
Reminders:
HW 9 due Monday 10pm
Midterm 3 a week today (Nov 12th)
Start making a 3rd formula card
Tuesdays lecture will be mainly for review
This is a task for Electron man!
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Circuits – Think like an electron
2. Wires: Not much to bump into – Low R.
Lose just a little bit of energy
1. Start: Lots of
energy
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Useful tip:
In questions, assume
connecting wires
have zero R and elose zero energy in
them, unless told
otherwise
• Circuits wired in series
• Circuits wired in parallel
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Rules: a) no electron deaths/births
b) no passing of electrons
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Circuits – Think like an electron
3. Filament
Lots to bump into
Higher R
(Like trudging
through mudpit)
Lose lots of energy.
c) electrons have energy
(high at start , low at end)
d) Material has resistance
(lets electrons pass easily or not)
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4. End: Exhausted! All energy
used up getting through course.
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See: resistor heating simulation:
http://phet.colorado.edu or class website
• Same current through connecting
wires and filament.
• Rfilament >> Rwires
• Almost all energy lost in filament.
Resistance is measure of how hard it is for electrons
to pass through an object … or how much stuff
they will run into.
Circuit language
Resistance (R) of a circuit element is measure of how hard
it is for electrons to pass through.
Units: Ohms (W)
Current (I) : charge per second flowing past a point in the
circuit
(= electrons per second × charge on electron)
Units : Amps (1 A = 1 C/s)
Voltage (difference) (V)
a) Across battery: Measure of EPE given to each e- as it passes through
battery. EPE given = eV. Related to pushing force on electrons in circuit
b) Across a resistor (wire, filament etc): Measure of EPE lost by each e- as
it passes through. EPE lost = eV.
Unless told otherwise voltage difference across connecting wire = 0.
Units: Volts (V)
Note: All quantities specific to
one component.
Ohm’s Law: V = IR
Don’t mix and match!
Resistance of component
Current through component
Voltage dropped across component
Vbatt
What if increase resistance (R) of filament… add
more stuff for e- to hit…
a. Rate at which electrons pass through filament
stays the same
b. Rate at which electrons pass through filament
decreases
c. Rate at which electrons pass through filament
increases
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11/3/2015
What if increase voltage difference across battery?
a. Rate at which electrons pass through filament stays
the same
b. Rate at which electrons pass through filament
decreases
c. Rate at which electrons pass through filament
increases
Vbatt
If the battery on the left has a voltage (difference) of 6V
and it is pushing a current of 1.5 A through the bulb, what
is the resistance of the bulb?
a)
b)
c)
d)
e)
9W
6W
4W
1.5 W
0W
1.5A
6V
Electrical Power
Power question
What is the electrical power used up by each component in circuit?
P = IV
Don’t mix and match!
Voltage dropped across component
Current through component
Electrical power dissipated (used up) in component
Also Ohm’s Law: V = IR
Substitute into power law to get different forms:
I have a 60W bulb plugged into the mains.
Assume that the mains supply is like a 120V battery
60W
What current flows through the bulb?
a) 120A
b) 60A
c) 0.5A
d) 2A
e) 7200A
120V
P = V2/R
- I = V/R
- Useful if you know V and R but not I (parallel circuits)
P = I2R
- V = IR
- Useful if you know I and R but not V (series circuits)
P = IV
- Useful if you know I and V but not R
Power question
I have a 60W bulb plugged into the mains.
Assume that the mains supply is like a 120V battery
Bulbs in series and parallel
Given 2 bulbs and a battery, there are 2 different ways that you
can connect the circuit
60W
What current flows through the bulb?
1.5 V
What is the resistance of the bulb filament?
a)
b)
c)
d)
e)
240 W
2W
0.5 W
30 W
Can’t determine
1.5 V
+
+
120V
Series
Parallel
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11/3/2015
Bulbs in series
Compare the brightness of identical bulbs.
How does bulb 1 compare to bulb 2?
a. brighter,
b. dimmer,
120 V
c. same
Electron man picture of bulb problem
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Electrons marching around in a line, pushing each other
through. Each electron passes through both bulbs.
Same number of electrons passing by per second at any
point on the circuit. So same current!
2
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Large
R
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Remember rules:
- No Passing
- No electron man deaths or
births on route
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Series circuit question - current
Large
R
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Series circuit question - voltage
case 1
All bulbs(R) and batteries (V) are identical.
How does current flow through circuit in case 1
compare to case 2?
a. Current in case 2 is 1/4 of current in case 1
b. Current in case 2 is 1/2 of current in case 1
c. Current in case 2 is same as current in case 1
d. Current in case 2 is 2 times current in case 1
e. Current in case 2 is 4 times current in case 1
+
case 2
120 V +
How big is voltage drop (difference) across
lefthand bulb (both bulbs identical)?
a. 60 V,
b. 120 V,
c. 240 V,
d. 0 V
?.?? V
+
+
*
Series circuit question - power
How will brightness of bulb 1 in circuit A compare to brightness of the
same bulb in the circuit B?
a. brighter A,
b. dimmer in A,
+
120 V
c. same
+
120 V
1
2
1
Bulbs in parallel
Compare the brightness of identical bulbs.
How does bulb 1 compare to bulb 2?
a. brighter,
b. dimmer,
c. same
+
120 V
1
2
Circuit B
Circuit A
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11/3/2015
Thinking like an electron – parallel circuits
Bulbs in parallel
No matter what the path, electrons start out with the same
energy from the battery and lose all their energy ( voltage)
by the time they return to outlet.
120 V
Vpath1 = Vpath2 = Vbattery
For bulbs wired in parallel:
• Same voltage across each bulb
• V1 = V2
• (Assumes Rwires ~ 0)
• Most useful form of power equation
is usually P = V2/R
If Rwires~ 0  Vpath1 = V1 = Vbattery
Vpath2 = V2 = Vbattery
lots of energy
at start. (Vbattery)
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V2
V1
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V1
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Path 1
+
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Path 2
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V2
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No energy left
• Bulbs have same R and same V across them
• Pin = V2/R is the same  brightness the same
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*
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Bulbs in parallel
Bulbs in parallel
B
A
+
120 V
+
120 V
1
1
How does size of IB compare to I1 ?
(Bulbs 1 and 2 are identical)
+
IB
1
a)
b)
c)
d)
2
120 V
I1
IB is three times I1
IB is twice I1
IB is half I1
IB is the same as I1
2
How does brightness of bulb 1 in circuit A compare to the same bulb in B?
a. brighter,
b. dimmer,
c. same
120 V
120 V
+
+
Bulb 2
Bulb 2
Bulb 3
Light bulbs wired in series.
If bulb 3 burns out, what happens to the current through bulb 2 ?
Bulb 3
Light bulbs wired in parallel.
If bulb 3 burns out, what happens to the current through bulb 2?
a. decreases, b. stays the same, c. increases
a. decreases, b. stays the same, c. increases
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11/3/2015
House wiring
When have desk lamp plugged into wall outlet and plug hair dryer into same
outlet, the two are:
a. in series (each electron flows through lamp then hair dryer, or vice-versa)
b. in parallel (each electron has to choose whether to go through the lamp or
the hair dryer, but never goes through both)
If bulbs A and B both have a resistance of 20 ohms, what is the total electrical
power converted in the bulbs?
a. 0.056 W,
b. 0.113 W,
c. 0.226 W
1.5 V
+
d. 30 W,
e. 60 W
A
B
Summary of some important ideas
1. Current is conserved (electrons don’t appear/disappear)
2. Change in V over circuit = V of battery, or energy source
3. V= I R (Ohm’s law)
- useful for whole circuit (R total, Vtotal, give I total)
- or individual component (e.g. Rbulb, Vbulb give I
…….Don’t mix and match.
4. P = I V = I (IR) = I2R
= (V/R)V = V2/R
bulb),
power dissipated by resistance R
5. Resistors in series:
Resistances add: Rtot = R1 + R2
Current through all resistors is the same
6. Resistors in parallel:
Voltage drop across parallel legs of circuit is same
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