Physics 102: Lecture 05
Circuits and Ohm’s Law
Physics 102: Lecture 5, Slide 1
Summary of Last Time
• Capacitors
– Physical C = κε0A/d C=Q/V
– Series 1/Ceq = 1/C1 + 1/C2
– Parallel Ceq = C1 + C2
– Energy U = 1/2 QV
S mmar of Toda
Summary Today
• Resistors
– Physical
Ph i l R = ρ L/A V=IR
V R
– Series Req = R1 + R2
– Parallel 1/Req = 1/R1 + 1/R2
– Power P = IV
Physics 102: Lecture 5, Slide 2
Electric Terminology
• Current: Moving Charges
– Sy
Symbol: I I
– Unit: Amp ≡ Coulomb/second
– Count number of charges which pass point/sec
– Direction of current is direction that + charge flows
• Power: Energy/Time
– Symbol: P
– Unit: W tt ≡ Joule/second
U it Watt J l / d=
Volt Coulomb/sec
– P = VI
60 W= 60 J/s
Physics 102: Lecture 5, Slide 3
Physical Resistor
• Resistance: Traveling through a resistor, electrons
bbump iinto things
hi hi h slows
which l h ddown.
them
R = ρ L /A Units: Ohms Ω I
– ρ: Resistivity: Density of scatterers A
– L: Length of resistor -
– A: Cross sectional area of resistor
L
• Ohms Law I = V/R
– Cause and effect (sort of like aa=F/m)
F/m)
• potential difference cause current to flow
• resistance regulate the amount of flow
– Double potential difference ⇒ double current
– I = (VA)/ (ρ L)
Physics 102: Lecture 5, Slide 4
CheckPoint 1.1
11
Two cylindrical resistors are made from the
same material. They are of equal length 1 2
but one has twice the diameter of the
other.
th
61% 1.
1 R1 > R2
R = ρ L /A
7% 2. R1 = R2
32% 3. R1 < R2
Physics 102: Lecture 5, Slide 5
Comparison:
C
Capacitors
it vs. Resistors
R i t
• Capacitors store energy as separated charge: U=QV/2
– Capacitance: ability to store separated charge:
C = κε0A/d
– Voltage drop determines charge: V=Q/C + --
++ -
• Resistors dissipate energy as power: P
P=VI
VI
– Resistance: how difficult it is for charges to get through:
R = ρ L /A
– Voltage drop determines current: V=IR
• Don’t mix capacitor and resistor equations!
Physics 102: Lecture 5, Slide 6
Simple Circuit
I
• Phet Visualization ε R
I
• Practice…
– Calculate I when ε=24 Volts and R = 8 Ω
– Ohm’s Law: V =IR
I = V/R = 3 Amps
Physics 102: Lecture 5, Slide 7
Resistors in Series
• One wire:
– Effectively adding lengths:
• Req=ρ(L1+L2)/A R
– Since R ∝ L add resistance:
2R
Req = R1 + R2 =
R
Physics 102: Lecture 5, Slide 8
Resistors in Series
• Resistors connected end-to-end:
– If charge goes through one resistor, it
must go through other.
I1 = I2 = Ieq
– Both have voltage drops: Req
I
V1 R1
V1 + V2 = Veq
V1+V2
I
V2 R2
I
Physics 102: Lecture 5, Slide 9
CheckPoint 2.1
Compare I1 the current through R1, with
I10 the current through R10. R1=1Ω
ε 0
R10=10Ω
13% 1. I1<I10
51% 2. I1=I10 “Since they are connected in series, the current is the same
for every resistor. If charge goes through one resistor, it
must go through other.”
36% 3. I1>I10
Note: I is the same everywhere in this circuit!
Physics 102: Lecture 5, Slide 10
ACT: Series Circuit
R1=1Ω
Compare V1 the voltage across R1, with ε
0
R10=10Ω
V10 the voltage across R10.
1. V1>V10 2. V1=V10 3. V1<V10
V1 = I1 R1 = I x 1
V10 = I10 R10 = I x 10
Physics 102: Lecture 5, Slide 11
Practice: R1=1Ω
Resistors in Series ε0
R2=10Ω
Calculate the voltage across each resistor if
the battery has potential ε0= 22 volts.
Simplify (R1 and R2 in series):
•R12 = R1 + R2 = 11 Ω
•V12 = V1 + V2 R12
= ε0 = 22 Volts ε0
•I12 = I1 = I2 = V12/R12 = 2 Amps
Expand:
p
•V1 = I1R1 = 2 x 1 = 2 Volts
R1=1Ω
•V2 = I2R2 = 2 x 10 = 20 Volts
ε0
Check: V1 + V2 = V12 ? R2=10Ω
Physics 102: Lecture 5, Slide 12
Resistors in Parallel
• Two wires:
– Effectively adding the Area
– Since R ∝ 1/A add 1/R:
1/Req = 1/R1 + 1/R2
Used in your house!
R R = R/2
Physics 102: Lecture 5, Slide 13
Resistors in Parallel
• Both ends of resistor are connected:
– Current is split between two wires:
I1 + I2 = Ieq
– Voltage is same across each:
V1 = V2 = Veq
I1+I2
I1+I2
I1 I2
V R1 R2 V Req V
Physics 102: Lecture 5, Slide 14
I1+I2
CheckPoint 3.1
What happens to the current through R2 when the switch is
closed?
23% • I
Increases
30% • Remains Same
47% • Decreases
Physics 102: Lecture 5, Slide 15
ACT: Parallel Circuit
What happens to the current through the battery when the
switch
it h is
i closed?
l d?
(A) Increases
(B) Remains Same
Ibattery = I2 + I3
((C)) Decreases
Physics 102: Lecture 5, Slide 16
Practice:
R i t
Resistors iin P
Parallel
ll l
ε R2 R3
Determine the current through the battery.
Volts R2 = 20 Ω and R3=30
Let ε = 60 Volts, 30 Ω.
Ω
Simplify: R2 and R3 are in parallel
1/R23 = 1/R2 + 1/R3 R23 = 12 Ω ε R23
V23 = V2 = V3 = 60 Volts
I23 = I2 + I3 = V23 /R23 = 5 Amps
Physics 102: Lecture 5, Slide 17
ACT /
CheckPoint
4142
4.1,4.2 1 2 3
R 2R R/2
Which
Whi h configuration
fi ti has
h the
th smallest
ll t resistance?
it ?
36% A. 1
5% B 2
B.
C. 3
59%
Which configuration has the largest resistance?
B 2
B. 70%
Physics 102: Lecture 5, Slide 18
R1
Try it!
ε R2 R3
Calculate current through each resistor.
R1 = 10 Ω,, R2 = 20 Ω,, R3 = 30 Ω,, ε = 44 V
Simplify: R2 and R3 are in parallel R1
1/R23 = 1/R2 + 1/R3 : R23 = 12 Ω
V23 = V2 = V3 ε R23
I23 = I2 + I3
Simplify: R1 and R23 are in series
R123 = R1 + R23 : R123 = 22 Ω
V123 = V1 + V23= ε ε R123
I123 = I1 = I23 = Ibattery : I123 = 44 V/22 Ω = 2 A
Power delivered by battery? P=IV = 2×44 = 88W
Physics 102: Lecture 5, Slide 19
Try it! (cont.)
ε R123
Calculate current through each resistor.
R1 = 10 Ω,, R2 = 20 Ω,, R3 = 30 Ω,, ε = 44 V
Expand: R1 and R23 are in series R1
R123 = R1 + R23 : I23 = 2 A
V123 = V1 + V23= ε : V23 = I23 R23 = 24 V
ε R23
I123 = I1 = I23 = Ibattery
Expand:
p R2 and R3 are in p
parallel R1
1/R23 = 1/R2 + 1/R3
V23 = V2 = V3 I2 = V2/R2 =24/20=1.2A ε R2 R3
I23 = I2 + I3
I3 = V3/R3 =24/30=0.8A
Physics 102: Lecture 5, Slide 20
Summaryy
Series Parallel
R1
R1 R2
R2
Each resistor on the Each resistor on a
Wiring same wire. different wire.
Different for each resistor. Same for each resistor.
Voltage Vtotal = V1 + V2 Vtotal = V1 = V2
Same for each resistor Different for each resistor
Current Itotal = I1 = I2 Itotal = I1 + I2
Increases Decreases
Resistance Req = R1 + R2 1/Req = 1/R1 + 1/R2
Physics 102: Lecture 5, Slide 21
