Chapter 19
A Microscopic View
of Electric Circuits
Current in a Circuit
A microscopic view of electric circuits:
• Are charges used up in a circuit?
• How is it possible to create and maintain a nonzero electric
field inside a wire?
• What is the role of the battery in a circuit?
In an electric circuit the system does not reach equilibrium!
Steady state and static equilibrium
Static equilibrium:
• no charges are moving
Steady state (Dynamic Equilibrium):
• charges are moving
• their velocities at any location do not change with time
• no change in the deposits of excess charge anywhere
Current in Different Parts of a Circuit
IB = IA in a steady
state circuit
We cannot get something for nothing!
What is used up in the light bulb?
Energy is transformed from one form to another
Electric field – accelerates electron
Friction – energy is lost to heat
Battery – chemical energy is used up
Closed circuit – energy losses to heat:
not an isolated system!
The Drude Model
E
Dp
Momentum principle:
= Fnet
Dt
p eEDt
Speed of the electron: v =
=
me
me
eE Dt
Average ‘drift’ speed: v =
me
Dp = Fnet Dt = eEDt
Dp = p - 0 = eEDt
Dt - average time between
collisions
The Drude Model
eE Dt
Average ‘drift’ speed: v =
me
Dt - average time between
collisions
For constant temperature v ~ E
v = uE ,
e
u=
Dt
me
u – mobility of an electron
Electron current: i = nAv
i = nAuE
Paul Drude
(1863 - 1906)
E and Drift Speed
In steady state current is the same everywhere in a series circuit.
Ethin
Ethick
i
i
What is the drift speed?
i = nAv
nAthin vthin = nAthick vthick
vthin
Athick
=
vthick
Athin
Note: density of electrons n cannot change if same metal
What is E?
v = uE
uEthin
Athick
=
uEthick
Athin
Ethin
Athick
=
Ethick
Athin
Question
i = nAv
1.5.n1 = n2
2.u1 = u2
1018
1 mm
2 mm
Every second 1018 electrons enter the thick wire.
How many electrons exit from the thin wire every second?
1018
A)
B) 1.5 x 1018
C) 2 x 1018
D) 4 x 1018
E) 12 x 1018
Question
i = nAv
v = uE
1.5.n1 = n2
2.u1 = u2
1018
1 mm
2 mm
What is the ratio of the electric field, E1/E2?
A)
B)
C)
D)
3:1
6:1
8:1
12:1
Direction of Electric Field in a Wire
E must be parallel to the wire
E is the same along the wire
Does current fill the wire?
Is E uniform across the wire?
B
C
D
A
A
B
C
D
DVABCDA = - ò E1 ×dl - ò E3 ×dl - ò E2 ×dl - ò E3 ×dl = 0
VAB
0
E1 = E2
VCD
0
Connecting a Circuit
When making the final connection in a circuit, feedback
forces a rapid rearrangement of the surface charges
leading to the steady state.
This period of adjustment before establishing the steady
state is called the initial transient.
The initial transient
Connecting a Circuit
1. Static equilibrium: nothing moving
(no current)
2. Initial transient: short-time process
leading to the steady state
3. Steady state: constant current
(nonzero)
Current at a Node
The current node rule
(Kirchhoff node or junction rule [law #1]):
i1 = i 2
i2 = i3 + i 4
In the steady state, the electron current entering a node in a
circuit is equal to the electron current leaving that node
(consequence of conservation of charge)
Gustav Robert Kirchhoff
(1824 - 1887)
Exercise
Write the node equation for this circuit.
What is the value of I2?
I1 + I4 = I 2 + I 3
I2 = I1 + I4 - I3 = 3A
What is the value of I2 if I4 is 1A?
I1 + I 4 = I 2 + I 3
I2 = I1 + I4 - I3 = -2A
1A
Charge conservation:
Ii > 0 for incoming
å Ii = 0 Ii < 0 for outgoing
i
Energy in a Circuit
Vwire = EL
Vbattery = ?
Energy conservation (the Kirchhoff loop rule [2nd law]):
V1 + V2 + V3 + … = 0
along any closed path in a circuit
V= U/q  energy per unit charge
Potential Difference Across the Battery
Coulomb force on each e
FC
non-Coulomb force on each e
FC
1. FC =eEC
EC =
e
2. FC =FNC
EC
DVbatt
FC s FNC s
= EC s =
=
e
e
Fully charged battery.
Energy input per unit charge
emf – electromotive force
The function of a battery is to produce and maintain a charge separation.
The emf is measured in Volts, but it is not a potential difference!
The emf is the energy input per unit charge.
chemical, nuclear, gravitational…
Field and Current in a Simple Circuit
Round-trip potential difference:
DVbatt + DVwire = 0
emf + ( - EL) = 0
E=
emf
L
emf
I = enAuE = enAu
L
We will neglect the battery’s internal resistance for the time being.
Field and Current in a Simple Circuit
Round-trip potential difference:
Path 1
DVbatt + DV1 + DV3 = 0
emf + ( - E1L1 ) + (- E3L3 ) = 0
Path 2
E1 L1 = E2 L2
DVbatt + DV2 + DV3 = 0
emf + ( - E2 L2 ) + (- E3L3 ) = 0
V Across Connecting Wires
The number or length of the
connecting wires has little effect on
the amount of current in the circuit.
DV
L
DV =
i
L
nAu
+ DVbattery = 0
i = nAuE = nAu
DVwires + DVfilament
DVbattery » emf
uwires >> ufilament  DVwires << DV filament
DV filament » (emf )
Work done by a battery goes mostly into energy dissipation in the
bulb (heat).
Twice the Length
Nichrome wire (resistive)
DV
i = nAuE = nAu
L
i2 L =
1
iL
2
Current is halved when increasing the length of the wire by a
factor of 2.
Doubling the Cross-Sectional Area
𝑖 = 𝑛𝐴𝑢𝐸
Loop: emf - EL = 0
𝐸=
𝑒𝑚𝑓
𝐿
Electron current in the wire increases
by a factor of two if the crosssectional area of the wire doubles.
Nichrome wire
Two Identical Light Bulbs in Series
i = nAuE = nAu
DV
L
Two identical light bulbs in
series are the same as one light
bulb with twice as long a
filament.
Identical light bulbs
The filament lengths add … 𝑅𝑒𝑞 = 𝑅1 + 𝑅2
Two Light Bulbs in Parallel
DV
i = nAuE = nAu
L
1. Path ABDFA:
-2 ( emf ) + EB L = 0
L … length of bulb filament
2 ( emf )
EB =
L
2. Path ACDFA:
-2 ( emf ) + EC L = 0
EC =
2 ( emf )
L
3. Path ABDCA:
EB L - EC L = 0
F
EB = EC
iB = iC
ibatt = 2iB
EB = EC
We can think of the two bulbs in parallel as equivalent to
increasing the cross-sectional area of one of the bulb filaments.
The filament areas add … 1/𝑅𝑒𝑞 = 1/𝑅1 + 1/𝑅2
Analysis of Circuits
The current node rule (Charge conservation)
Kirchhoff node or junction rule [1st law]:
In the steady state, the electron current entering a node in a
circuit is equal to the electron current leaving that node
Electron current: i = nAuE
Conventional current: I = |q|nAuE
The loop rule (Energy conservation)
Kirchhoff loop rule [2nd law]:
V1 + V2 + V3 + … = 0
along any closed path in a circuit
V= U/q  energy per unit charge