Electromagnetism
Physics 15b
Lecture #9
DC Circuits
Purcell 4.7–4.11
What We Did Last Time
I=
Define current I and current density J
∂ρ

Charge conservation divJ = −
∂t

J ⋅ da
J = ∑ nk qk uk
k
J = σ E (micro) V = RI (macro)
For constant cross section A, R =
L
σA
Physics of electrical conduction

S
or = 0 in steady current
Carrier densities nk, velocities uk
Ohm’s Law
∫
Assuming that carrier (=electron) motion is
dominated by thermal fluctuation
σ=
ne e 2τ
me
1
Today’s Goals
One more resistance problem

How to go between micro (resistivity)
and macro (resistance)
Power dissipation in a resistor
Discuss direct current (DC) circuits

Made of resistors, capacitors and
electromotive forces
Resistor arithmetic

Many of you already know
Kirchhoff’s rules

Simple and straightforward rules that
help you to solve DC circuit problems
systematically
Gustav Kirchhoff (1824–1887)
Cylindrical Resistor
A material of resistivity ρ is filled between two highlyconductive tubes of radii a and b (a < b), length L

Apply potential difference V between tubes
E and J are both pointing out

At radius r, J(r ) = σ E(r ) =

b
a
1
E(r )r̂
ρ
J
r
Integrate J over a cylinder at r
I=
∫
J(r ) ⋅ da = 2π rLJ(r )
E(r ) = ρ J(r ) =
cylin der
ρI
r̂
2π rL
Integrate E from r = a to b
V=
∫
b
a
E(r ) ⋅ dr =
⎛ b⎞
ρI b dr
ρI
=
ln ⎜ ⎟
∫
a
2π L
r
2π L ⎝ a ⎠
R=
⎛ b⎞
V
ρ
=
ln ⎜ ⎟
I 2π L ⎝ a ⎠
2
Power Dissipation
Current I flows through a piece of resistive wire
Every t seconds, charge Q = It
moves across potential difference V
  They lose energy U = QV = IVt
  What happens to the energy?

q
q
I
q
V
I=
V
R
Current I flowing downstream across V dissipates power
P = IV = I 2R
and turn it into heat

Unit: erg/sec in CGS, joule/sec = watt (W) in SI
Next: We need a source of energy to keep it going
Electromotive Force
Batteries supply “electricity”
Connect it to a resistor and we’ll find V and I
  Connect it to a capacitor and we’ll find V and Q

I
R
+
V
−
+Q
C
−Q
+
V
−
A battery is a potential-difference generator
We call a battery’s ability to produce a potential difference an
electromotive force, or emf
Watch out! It’s not a force
  Unit: statvolt in CGS, volt in SI

3
What is an EMF
Inside a battery, chemical reactions move the charges from
one contact to another

It’s like a pump pushing water up
−
+
−
+
h
−
R
+
V
−
I
+
Battery converts chemical energy
to potential energy of charges
Pump converts kinetic energy
to potential energy of water
Internal Resistance
Connect a battery with an emf E to a resistor R
E
R

The current should be I =

True only for an ideal battery
I
R
+ E!
V
Real batteries have internal resistance
+
Ri
−
E!
Internal resistance

−
Ideal battery
V measured outside the battery is
Ri
I
R
V = E − IRi < E
V
+ E!
−
Loss due to internal resistance
4
Measuring E and Ri
Internal loss, I·Ri, is proportional to the current
We can measure V with no current and get E!
  This is open-circuit voltage
  We can measure V with current and find out Ri

Ri
E!
V
R
+ E!
Ri
I
−
+ E!
I=
E
R + Ri
V=
ER
R + Ri
−
Battery testers do this to determine if Ri is too large

“Dead” batteries have the same emf E as the fresh ones, but larger
internal resistance Ri
Series/Parallel Resistors
Series
Parallel
R1
R2
I
+
−
Current I through R1 and R2
are the same
  Potential differences across
R1 and R2 are V1 = IR1 and
V2 = IR2
  Total potential V must be

V = V1 +V2 = IR1 + IR2 ≡ IRequiv.
Requiv. = R1 + R2
R1
V
R2
+
−
V
Potential V across R1 and R2
are the same
  Currents through R1 and R2
are I1 = V/R1 and I2 = V/R2
  Combined, they draw

I = I1 + I2 =
1
Requiv.
V
V
V
+
≡
R1 R2 Requiv.
=
1
1
+
R1 R2
5
Resistor Arithmetic
For arbitrary number of resistors
R1
R2
R3
…
R1
Rn
R2
Requiv. = ∑ Ri
1
i
Requiv.
Generally, any 2-terminal network
of resistors is equivalent to a single resistor

It may be easy
or not so easy
R2
R1
R3

R3
Rn
…
=∑
1
Ri
i
R1
R3
R2
R4
Not so easy ones need a strategy
Kirchhoff’s Rules
#1) At any point in a circuit, the sum of the incoming current
equals to the sum of the outgoing current
∑I
in
= ∑ Iout
Junction Rule
#2) Around any loop in a circuit, the sum of the potential
differences is zero
V = 0 Loop Rule
∑
loop
These are restatements of what we know perfectly well
  Electric charge is conserved
d
ρ dv = 0
∇⋅J = 0
 Current is divergent free
dt ∫V

Electric potential is conservative
 Potential is curl free
∇×E = 0
∫
C
E ⋅ds = 0
6
Using Kirchhoff
Let’s do a simple example: Find the current through R3
Step 1: Define currents


One for each “branch” of the circuit
Direction doesn’t matter – Pick one
I1
R1
+
E1
−
Step 2: Apply the junction rule
R3
Junction A  I1 + I2 = I3
  Junction B  I3 = I1 + I2
A

Step 3: Apply the loop rule
R2
Top loop
 E1 − I1R1 − I3R3 = 0
  Bottom loop  −E2 − I2R2 − I3R3 = 0

B
I3
+
I2
−
E2
What’s this minus sign doing here?
Loop Rule Subtlety
To apply the loop rule, you must decide which direction to
go around each loop

Either will work, just pick one
I1
As you go around the loop,
An emf counts as +E if you pass
through it from − to +
  A resistor counts as −IR if the
current is flowing the direction you
are going

−E
+
E1
−
R3
A
R2
B
I3
+
I2
−
E2
−E2 − I2R2 − I3R3 = 0
−
+
−
+E
+
If you are going this way …
R1
−IR
+IR
7
Back to the Problem
We’ve found so far

Junctions:
I1 + I2 = I3

Top loop:
E1 − I1R1 − I3R3 = 0

Bottom loop: −E2 − I2R2 − I3R3 = 0
I1
R1
We introduced 3 unknown currents
and found 3 equations

Solve them:
I3 =
E1
−
R3
A
(R2 + R3 )E1 + R3E2
I1 =
R1R2 + R1R3 + R2R3
I2 = −
+
B
R2
R3E1 + (R1 + R3 )E2
R1R2 + R1R3 + R2R3
I3
+
I2
−
E2
R2E1 − R1E2
R1R2 + R1R3 + R2R3
Kirchhoff Strategy
+E
Step 2: Apply the junction rule
Step 3: Apply the loop rule
Pick the direction of the loop
  Pay attention to the polarity

loop
− IR
−
−
+
One for each “branch” of the circuit
  Pick the direction of the current

+
Step 1: Define currents
−E
+ IR
You’ll have just the right number of equations to solve for
all the currents you defined in Step 1

The rest is straightforward (it may be tedious)
8
Thévenin/Norton Equivalences
Any 2-terminal circuit of emfs and resistors can be replaced
with an emf and a resistor in series, or a current source and
a resistor in parallel
Thévenin
Norton
RTh
INo!
+
−
+
ETh!
RNo
−
They cannot be electrically distinguished from outside
  They have the same I-V relation on the terminals

Symbol for a
current source
The equivalence theorems assume linearity (including
reversibility) of the emfs and resistors
Finding Equivalences
Two parameters  Two points on the I-V relation
Open-circuit voltage VO: potential difference between
the terminals with nothing connected
  Short-circuit current IS : current between the
terminals when they are “shorted”

VO = ETh , IS = ETh RTh
for Thévenin
VO = INo RNo , IS = INo
for Norton
VTh = VO , INo = IS , RTh = RNo = VO IS
Let’s do an example
RTh
+
−
ETh!
+
INo!
RNo
−
9
Example Thévenin/Norton
Put the circuit we looked at before in a “black” box

Open circuit : we have already solved
for I1, I2, and I3
Short circuit : connecting A and B
means adding 0Ω in parallel with R3
  Effectively this makes R3 = 0
  I3 will become IS

IS = I3 R
3
We find: Req =
I1
R3 (R2E1 − R1E2 )
R1R2 + R1R3 + R2R3
V0 = R3I3 =
=0
Vo
IS
R1
=
−
E1
R3
A
R E − R1E2
= 2 1
R1R2
+
R2
B
I3
+
I2
−
E2
R1R2R3
1
=
R1R2 + R1R3 + R2R3 1 R1 + 1 R2 + 1 R3
Equivalent Resistance
Thévenin/Norton equivalent resistance turns out to be
Req =
1
= R1  R2  R3
1 R1 + 1 R2 + 1 R3
One can find Req by replacing all
emfs with 0Ω and calculating the
A
total resistance
Consider the differential I-V relation

For a resistor

For an emf
dV d(RI)
=
=R
dI
dI
dV dE
=
=0
dI
dI
R1
0Ω
R3
R2
B
0Ω
i.e., an emf’s differential resistance is 0
10
Summary
Power dissipation in resitors P = IV = I R
Electromotive forces (emfs)
2

Batteries are made of an emf and an internal resistance
Resistor arithmetic Rseries = ∑ R1
i
1
Rparallel
=∑
i
1
Ri
Kirchhoff’s rules
∑I
in

= ∑ Iout
Junction Rule
∑V = 0
Loop rule requires attention to the polarity
Tévenin and Norton equivalences
Open-circuit voltage and short-circuit current
  Req can be evaluated by emf = 0Ω

Loop Rule
loop
VTh = VO
INo = IS
RTh = RNo = VO IS
11