ELECTRIC CURRENT
The 3 quantities describing a DC circuit are:
1) Current (I)
2) Electric Potential (V)
3) Conductance (g).
Definitions:
Current = flow of charge, Q, through a cross-sectional area in time, t.
I = Q/t
SI unit = Ampere (A) = (1 Coulomb/second).
Motion of electrons (negative!) actually produces the current. For e.g. in a
copper wire:
Copper atomic cores
Direction of Conventional Current
Free Electrons
A negative current or a positive current moving in the opposite direction are
equal mathematically. Therefore, we treat all currents as the positive flow
of charge in problem solving.
Electric Potential - symbol, VAB, if the work (Energy) in moving a charge, q,
from a point A to a point B is WAB. Then,
WAB = qVAB
and the electric potential is
VAB = WAB/q
SI units = Joules/Coulomb = Volt (V). Electric potential is often called the
voltage.
Conductance current per applied voltage - depends on conducting material.
Consider a voltage, VAB, across a material with conductance, g:
g
A
B
I
VAB
If g is a constant above quantities obey
I = gVAB .
UNITS:
Ohm's Law
Amperes /Volt = Siemens (S).
Resistance (unit: Ω) is the inverse of conductance, R = 1/g. Then, V = IR
(alternate form of Ohm’s law)
CALCULATING CONDUCTANCE:
Dependant on the dimensions and electric properties of material. Consider a
conductor with cross-sectional area, A and length L;
L
current
A
Its conductance is ,
g = σA/L ,
where σ is the conductivity of the material.
DC CIRCUITS:
•
•
Current varies sinusoidally on power grids - Alternating Current (AC)
We study constant current - Direct Current (DC).
g
A simple DC circuit
I
V
Kirchoff's rules
1. The net voltage around a closed loop is zero.
2. Current entering a point is equal to current leaving it
Other useful observations:
•
•
points connected by a PERFECT conductor are at the same electric
potential.
Conductors arranged in parallel and series have specific rules to
determine their effective conductance.
Parallel Network:
g1
I1
g2
I
A
I
I2
g3
I3
V
The equivalent conductance, geq, such that I = geqV, for the 3 conductors
above is found using Kirchoff's second rule at, A,
Current in
I = I1 + I2 + I3
Current out
NB: all 3 conductors have the same voltage across them, V, (parallel)
therefore;
I1 = g1V
I2 = g2V
I3 = g3V
∴ I = g1V + g2V + g3V
= (g1 + g2 + g3)V
therefore
geq = g1 + g2 + g3
SERIES NETWORK:
g1
I
g2
g3
A
B
V1
V2
VAB
Rearranging Ohm’s law: I/geq = VAB. Notice…
V3
1. Each conductor now experiences a different V.
2. VAB = V1+V2+V3
3. Current out = current in, so each conductor has the same current, I.
Ohm’s law for each conductor;
I/g1 = V1
I/g2 = V2
I/g3 = V3
∴ I/geq = I/g1 + I/g2 + I/g3 .
Dividing by I on both sides, we find the equivalent conductance,
1
geq
=
1 1 1
+ +
g g g .
1 2 3
EXAMPLE 1: Conductors in Parallel
Calculate geq for the following network of conductors:
g1 = 2.0 × 10-3 S
g2 = 3.0 × 10-3 S
g3 = 6.0 × 10-3 S
As shown earlier, geq = g1 + g2 + g3. Therefore, geq = (2.0 + 3.0 +6.0) × 10-3 =
11.0 × 10-3. That is, geq = 1.1 × 10-2 S.
EXAMPLE 2: Conductors in Series
For the same three conductors in a series arrangement:
g1
g2
Since 1/geq = 1/g1 + 1/g2 + 1/g3
then
1/geq = 1/(2.0 × 10-3) + 1/(3.0 × 10-3) + 1/(6.0 × 10-3)
= 500 +333.33 + 166.67 = 1000 S-1 (or Ω).
Therefore:
geq = 1/1000 = 1.0 × 10-3 S.
g3
EXAMPLE 3: Series/Parallel Conductor Network
Calculate geq for the following combination of conductors:
g1 = 2.0 × 10-3 S
g3 = 1.0 × 10-3 S
g2 = 4.0 × 10-3 S
g4 = 6.0 × 10-3 S
The network can be simplified by noting that its conductance is equal to:
geq1 = g1 + g2
geq2 = g3 + g4
using the parallel conductors rule. We can now use the series rule to find
geq. i.e.)
1/geq = 1/geq1 + 1/geq2
Subbing in the values for geq1 and geq2
1/geq = 1/(g1 + g2) + 1/(g3+g4)
Substitution of the given values for the conductances gives
1/geq = 166.67 + 142.86 = 309.53 S-1.
Therefore, geq = 1/309.53 S or…
geq = 3.23 × 10-3 S.
EXAMPLE 4: Series, Parallel or What?
Calculate geq for the following network. All the conductors are 1.0 S.
Wires are assumed to be perfect conductors and can be re-arranged as long
as the same paths are possible in the new network as the old.
This is series-parallel arrangement. Using a similar to example 3 to solve for
geq, we find that
geq = 1.2 S.
EXAMPLE 5: Conductance of a cylinder
Calculate the conductance of the cylinder below:
2.00 cm
0.500 mm
σ = 1.50 × 10-2 S/cm
Using our formula from before: g = σA/L
g =
(
)
-2 

2
1.50×10  ×π × 0.0250


2.00
=1.47×10− 5 S