Today: Ohm’s Law
• Checkpoints Review
64%
I1 / I2=2
67%
V2 = V3 < V
I2 = I3
62%
V1 > V2 for same I
V1 < V2 for same I
“I'm just curious to see if this quote gets on the lecture
slides;” -Ian
V1 = 2V2 = V
• Capacitors:
Devices
Purpose: store charge (energy).
» Several geometries calculated to determine C
» Modified to account for dielectrics: C = C0
» Effective capacitance of series or parallel combinations
• Batteries (Voltage sources, seats of emf):
Purpose: provide a constant potential difference
between 2 points.
» Cannot calculate from first principles
» electrical  chemical energy conversion.
» Non-ideal batteries will be dealt with in terms
of an "internal resistance".
+ OR
+ V -
Devices
• Resistors:
Purpose: limit current in a circuit.
Note:
dq
I
dt
UNIT: Ampere = A = C/s
• Resistance will depend on geometry and a material’s “resistivity”
Current Density
J=I/A
Note: Our CONVENTION for current flow is + to – as dashed line shows
as shown nicely in the SP animation, the electrons “flow” opposite
Free charges under E field accelerate
Current is the amount of charge per unit time moving
thorough a given section of wire:
dQ
I
dt
Expect electrons in a metal to accelerate and yield a current
that gets larger with time?

e 
v   Et
m
No: electrons have a random velocity that is of
order 106 m/s. The electrons suffer collisions and
change their directions. The field changes the
magnitude only slightly.
Model currents in metals under electric fields: the Drude model
conventional current opposite the drift
velocity of e’s (in the direction of E).

e 
vd   E
m
Typical vd’s are 10-4 m/s.



nq 2
J   vd  qn vd 
m



E


J  E
conductivity
In a wire of area A and length d:
Ohm’s law
JA  I  A
1 d
Ed
  / R with R 
d
 A
I   / R
Solving circuits
Useful rules:
1) Sum of all potential differences add
up to zero in a closed loop.
 
i
0
closed loop
2) Sum of currents coming to a junction
should add up to zero.
i
i
junction
0
Notice the arrows
are all toward the
joint. If you have a
circuit with currents
coming out, you
need to flip their
signs.
Example 1: Consider the circuit
at right. Given E1, R1, R2, R3,
calculate i1, i2, i3.
Recipe 1: First draw the signs of  for the given choices of currents
Recipe 2: Next draw loops and
write the eqs.
 
i
 E1  i1 R1  i 2 R2  0
 
 E1  i1 R1  i 3 R3  0
loop 1
i
loop 2
i1  i 2  i 3  0
These two circuits are equivalent
with
Proof:
Req  R2  R3
The equivalent resistance is higher
than either.
E1  i 1 R1  i 1 R2  i 1 Req
These two circuits are equivalent
with
Proof:
1
1
1


Req R2 R3
The equivalent resistance is lower
than either.
i1  i 2  i 3
  


Req R1 R2
The battery has zero internal
resistance. The circuit
elements have the following
values: E = 20 V, R0=2/3 ,
R1=4 , R2=4 , R3=3 ,
and R4=1 .
Is the magnitude of the current I larger than,
equal to, or smaller than |IBG+ICF|
A. Larger.
B. Equal.
C. Smaller.
Ohm's Law
I
• Demo:
R
I
• Vary applied voltage V.
• Measure current I
V
• Does ratio (V/I) remain
constant?
R
V
V
I
slope = R
An empirical finding
I
Resistivity
“Current density is initially a little tricky, I missed that it
was per unit area and not per unit volume, but I think I got
the right answer nevertheless..” --Clark
Property of bulk matter related to
resistance: resistivity 
E
 
J
E = electric field
J = current density
For uniform case:

I
J
A
E
J
A
L
& V  EL
I
 L 
V  EL  JL   L  I  
A
 A

R
The resistance, R, is a property of the device;
The resistivity, , is a property of the material
eg, for a copper wire,  ~ 10-8 -m, 1 mm radius, 1 m long,  R  0.01
L
A
prelecture
L
R
A
checkpoint
The SAME amount of current I passes through three different resistors.
• R2 has twice the cross-sectional area and the same length as R1,
• R3 is three times as long as R1 but has same cross-sectional area as
R1.
In which case is the CURRENT DENSITY through the resistor
the smallest?
J
I
A
Clicker
• Two cylindrical resistors, R1 and R2, are made of identical material. R2
has twice the length of R1 but half the radius of R1.
– These resistors are then connected to a battery V as shown:
I1
I2
V
– What is the relation between I1, the current flowing in R1 , and I2 , the
current flowing in R2?
(a) I1 = I2
(b) I1 = 2I2 (c) I1 = 4I2
The resistivity  is the same
Resistances are:
(d) I1 = 8I2
2 L1
L
L
 8 1  8R1
R2   2  
( A1 / 4)
A2
A1
The resistors have the same voltage across them; therefore
I2 
V
V
1

 I1
R 2 8 R1 8
(brief) Electric
Power
• Suppose a charge dq moves through
potential difference V.
– Its Potential Energy changes by dU = dq V
– The Rate of Energy Change (gain or loss) is:
dU dqV

 IV
P
dt
dt
– Applies to any electrical device that can deliver, store
or use electrical energy.
– Units = J / s  Watts (W)
– Applies at any instant of time, but also can be
averaged over a time interval, “the average power”
CLICKER
How is it that a Constant E- Field Produces
Constant Velocity of Moving Charges?
A.
B.
C.
D.
E.
Constant Force Usually Means Constant Velocity.
Newton's Laws Do Not Apply to Charged Particles.
Electrons Accelerate Randomly, So Average Acceleration = 0
"Drag" Force Exists, So the Net Force on Each Charge = 0
None of the Above.
Resistors in Series
a
I
R1
The Voltage “drops”:
Va  Vb  IR1
Vb  Vc  IR 2
b
Va  Vc  I(R1  R 2 )
Whenever devices are in SERIES, the
current is the same through both !
R2
a
c
This reduces the circuit to:
Reffective
Hence:
R effective  (R1  R 2 )
c
Resistors in Parallel
• Devices in parallel have the same voltage drop V
a
• But current through R1 is not I
– Call it I1.
– Similarly, current through R2 is I2.
• Current is conserved at junction!
I
I1
V


1
1
1


Req R1 R2
R2
R1
I
d
I =I1+I2
V V V


R R1 R 2
I2
I
a
V
R
d
I
Analog: Water Flow in Pipes
Current is like flow (in gallons/min)
Voltage is like pressure
Flow = Pressure diff / impedance
I = V/R
Calculation with Clickers
R2
R1
V
In the circuit shown: V  18V, R1  1R2  2R3  3and R4  4
R3
R4
What is V2, the voltage across R2?
Combine Resistances: R1 and R2 are connected:
A) in series B) in parallel C) neither in series nor in parallel
Parallel Combination Series Combination Ra
Ra
Rb
Parallel: Can make a loop that contains only those two resistors
Rb
Series : Every loop with resistor 1 also has resistor 2.
Shout-out “clicker”
R2
R1
In the circuit shown: V  18V, R1  1R2  2R3  3and R4  4
V
R3
Which is the correct equivalent circuit?
R4
R2 and R4 are connected in series R24which is connected in parallel with R3
Redraw using the equivalent resistor R24  series combination of R2 and R4. R1
V
R1
R3
R24
V
R1
R3
V
R3
R24
R24
(A) (B) (C) Resume Calculation and Click with Numbers
R1
V
R3
R24
In the circuit shown: V  18V, R1  1R2  2R3  3and R4  4
What is V2, the voltage across R2?
Combine Resistances: R2 and R4 are connected in series  R24
R3 and R24 are connected in parallel  R234
What is the value of R234?
A) R234  1  B) R234  2  C) R234  4 
R2 and R4 in series
1Rparallel1Ra  1Rb
D) R234  6 
R24  R2  R4  2  4  6
1R234 13 1636-1
R234  2 
Calculation
R1
V
I1  I234
RR234
234
In the circuit shown: V  18V, R1  1R2  2R3  3and R4  4
What is V2, the voltage across R2?
R1 and R234 are in series. R1234  1  2  3 
Our next task is to calculate the total current in the circuit
Ohm’s Law tells us: I1234  VR1234
V
 I1234
R1234
18 3
 6 Amps
Shout-out Clicker … Calculation
In the circuit shown: V  18V, R1  1R2  2R3  3and R4  4
V
 I1234
R1234
What is V2, the voltage across R2?
R1
I234  I1234 Since R1 in series with R234
a
V
I1  I234
V234  I234 R234
RR234
234
6x2
 12 Volts
b
What is Vab, the voltage across R234 ?
A) Vab  1 V
B) Vab  2 V
I1234  6 Amps
C) Vab  9 V
D) Vab  12 V
E) Vab  16 V
More shout outs …Calculation
R1
R1
V
V
R3
R234
R24
Which of the following are true?
A) V234  V24
B) I234  I24
C) Both A+B
R3 and R24 were combined in parallel to get R234
Ohm’s Law
I24  V24 / R24
 12 / 6
 2 Amps
D) None
V  18V
R1  1
R2  2
R3  3
R4  4
R24  6
R234  2
I1234  6 Amps
I234  6 Amps
V234  12V
What is V2?
Voltages are same!
And one more … Calculation
R1
I1234
V
R1
I24
R2
V
R3
R3
R24
R4
Which of the following are true?
A) V24  V2
B) I24  I2
C) Both A+B D) None
R2 and R4 where combined in series to get R24
Ohm’s Law
The Problem Can Now Be Solved!
V2  I2 R2
 2x2
 4 Volts!
V  18V
R1  1
R2  2
R3  3
R4  4
R24  6
R234  2
I1234  6 Amps
I234  6 Amps
V234  12V
V24  12V
I24  2 Amps
What is V2?
Currents are same!
And Some Quick Follow-Ups
I1
R1
I2
R2
R1
I3
V

R3
a
V
R234
R4
b
What is I3 ? A) I3  2 A
B) I3  3 A
V3 = V234 = 12V
C) I3  4 A
I3  V3/R3  12V/3  4A
V  18V
R1  1
R2  2
R3  3
R4  4
R24  6
R234  2
V234= 12V
V2 = 4V
I1234 = 6 Amps
What is I1 ? We know I1  I1234  6 A
NOTE: I2  V2R2  42  2 A
I1  I2  I3
Make Sense?
Another (intuitive) way... In Appendix
R1
Consider two cylindrical resistors with
lengths L1 and L2
R1  
L1
A
L1
V
L2
L
R2   2
A
R2
Put them together, end to end to make a longer one...
R effective  
L1  L 2
 R1  R 2
A
R  R1  R 2
Another (intuitive) way...In Appendix
Consider two cylindrical resistors with
cross-sectional areas A1 and A2
R1  
A1
L
A1
V
R1
L
R2  
A2
Put them together, side by side … to make one “fatter”one,
R effective 
L
A1  A 2 


1
R effective
1
1
1


R R1 R 2

A2
A1 A 2
1
1



L L R 1 R 2
R2