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Physics 212
Lecture 9
Today's Concept:
Electric Current
Ohm’s Law & resistors
Resistors in circuits
Power in circuits
Physics 212 Lecture 9, Slide 1
Main Point 1
First, we defined the electric current in a conductor as the amount of charge that
passes through a cross-section of the conductor per unit time (i.e., I = dq/dt). We
developed a microscopic view of current in which the charge carriers are
dissociated electrons in the conductor that move randomly at high speeds (~10^5
m/s), but when a potential difference is introduced across the conductor, the
resultant electric field gives these electrons a non-zero average velocity which is
Physics 212 Lecture 9, Slide 2
the source of the electric current.
Main Point 2
Second, we introduced Ohm’s law that states that, for a wide range of materials,
over a wide range of field strengths, the current density is proportional to the
electric field that gives rise to it. The constant of proportionality is called the
conductivity of the material. We used this constant to characterize the resistance
of a resistor, obtaining the more common form of Ohm’s law in which the voltage
across the resistor is equal to the product of the resistance and the current that
Physics 212 Lecture 9, Slide 3
flows through the resistor.
Main Point 3
Third, we obtained expressions for the equivalent resistance of two resistors
connected either in series or in parallel. Namely, the equivalent resistance of two
resistors in series is simply the sum of the individual resistances, while the inverse
of the equivalent resistance of two resistors connected in parallel is equal to the
sum of the individual inverse resistances.
Physics 212 Lecture 9, Slide 4
Main Point 4
Fourth, we determined that the power, the time rate of change of the energy of a
circuit component is always equal to the product of the voltage drop across the
component and the current that flows through the component. In a simple circuit
composed of a single resistor connected to the terminals of the battery, we found
that all of the energy that is supplied by the battery is dissipated as heat in the
Physics 212 Lecture 9, Slide 5
resistor.
I
s
A
L
V
J=sE
same as
I = V/R
resistivity – high for bad conductors.
where R =
L = rL
sA
A
Conductivity – high for good conductors.
Physics 212 Lecture 9, Slide 6
Physics 212 Lecture 9, Slide 7
Checkpoint 1a
Checkpoint 1b
Same current through
both resistors
Compare voltages
across resistors
Physics 212 Lecture 9, Slide 8
Physics 212 Lecture 9, Slide 9
Checkpoint 3
The SAME amount of current I passes through three different resistors. R2 has twice the crosssectional area and the same length as R1, and R3 is three times as long as R1 but has the same
cross-sectional area as R1.
In which case is the CURRENT DENSITY through the resistor the smallest?
A. Case 1
B. Case 2
C. Case 3
Physics 212 Lecture 9, Slide 10
Physics 212 Lecture 9, Slide 11
Three resistors are connected to a battery with emf V as
shown. The resistances of the resistors are all the same,
i.e. R1= R2 = R3 = R.
Checkpoint 2a
Compare the current through R2 with the current through R3:
A. I2 > I3
B. I2 = I3
C. I2 < I3
Physics 212 Lecture 9, Slide 12
Physics 212 Lecture 9, Slide 13
Checkpoint 2b
R 1 = R 2 = R3 = R
A. I1/I2 = 1/2
B. I1/I2 = 1/3
C. I1/I2 = 1
D. I1/I2 = 2
E. I1/I2 = 3
Physics 212 Lecture 9, Slide 14
Physics 212 Lecture 9, Slide 15
Checkpoint 2c
R 1 = R 2 = R3 = R
Compare the voltage
across R2 with the
voltage across R3
A
V2 > V3
B
V2 = V3 = V
C
V2 = V3 < V
D
V2 < V3
Physics 212 Lecture 9, Slide 16
Physics 212 Lecture 9, Slide 17
Checkpoint 2d
R 1 = R 2 = R3 = R
Compare the voltage across
R1 with the voltage across R2
A
V1 = V2 = V
B
V1 = ½ V2 = V
C
V1 = 2V2 = V
D
V1 = ½ V2 = 1/5 V
E
V1 = ½ V2 = ½ V
Physics 212 Lecture 9, Slide 18
Physics 212 Lecture 9, Slide 19
Resistor Summary
Series
Parallel
R1
R1
R2
R2
Wiring
Each resistor on the
same wire.
Each resistor on a
different wire.
Voltage
Different for each
resistor.
Vtotal = V1 + V2
Same for each
resistor.
Vtotal = V1 = V2
Current
Same for each resistor
Itotal = I1 = I2
Different for each
resistor
Itotal = I1 + I2
Decreases
1/Req = 1/R1 + 1/R2
Resistance
Increases
Req = R1 + R2
Physics 212 Lecture 9, Slide 20
R2
R1
V
Calculation
In the circuit shown: V = 18V,
R1 = 1W, R2 = 2W, R3 = 3W, and R4 = 4W.
R3
R4
What is V2, the voltage across R2?
• Conceptual Analysis:
–
–
Ohm’s Law: when current I flows through resistance R, the potential
drop V is given by: V = IR.
Resistances are combined in series and parallel combinations
• Rseries = Ra + Rb
• (1/Rparallel) = (1/Ra) + (1/Rb)
• Strategic Analysis
–
–
–
Combine resistances to form equivalent resistances
Evaluate voltages or currents from Ohm’s Law
Expand circuit back using knowledge of voltages and currents
Physics 212 Lecture 9, Slide 21
Physics 212 Lecture 9, Slide 22