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23/05/31
Chapter 17
Direct Currents
Electric Current
The current is the rate at which charge flows
through this surface.
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average current
If ΔQ is the amount of charge that passes through this area
in a time interval Δ t, the average current Iav is equal to the
charge that passes through A per unit time:
instantaneous current
If the rate at which charge flows varies in time, then
the current varies in time; we define the instantaneous
current I as the differential limit of average current:
The SI unit of current is the ampere (A):
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Microscopic Model of Current
The total charge in the wire ΔQ= env = enAl
Δt = l/ѵ
So the magnitude of the current is
I
Q enlA

 enAv
l
t
v
The current is the product of the electronic charge(e),the
density of conduction electrons(n),the area (A),and the
average drift velocity(ѵ).
Example:
Consider positive and negative charges moving horizontally
through the four regions shown in Figure. Rank the current in
these four regions, from lowest to highest.
d, b=c ,a.
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Example Drift Speed in a Copper Wire
The 12-gauge copper wire in a typical residential building has a cross sectional
area of 3.31 x 10-6 m2. If it carries current of 10.0 A, what is the drift speed of
the electrons? Assume that each copper atom contributes one free electron
to the current. The density of copper is 8.95 g/cm3.
Avogadro’s number of atoms (6.02 X 1023)
The molar mass of copper is 63.5 g/mol
Resistance
The resistance as the ratio of the potential difference
across a conductor to the current in the conductor:
Material with constant resistance are said to obey
ohm’s law and are called ohmic conductors.
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Resistance
The resistance has SI units of volts per ampere. One
volt per ampere is defined to be one ohm (Ω):
Resistivity
R
l
A
R
l
A
where ρ has the units ohm.meters (Ω.m).
conductivity
The inverse of resistivity is conductivity σ:
where σ has the units (Ω.m)-1.
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Example The Resistance of a Conductor
Calculate the resistance of an aluminum cylinder that has a
length of 10.0 cm and a cross-sectional area of 2.00 x 10-4 m2.
Repeat the calculation for a cylinder of the same dimensions
and made of glass having a resistivity of 3.0 x 1010 Ω.m.
Example The Resistance of Nichrome Wire
(A) Calculate the resistance per unit length of a 22-gauge
Nichrome wire, which has a radius of 0.321 mm.
(B) If a potential difference of 10 V is maintained across a 1.0m length of the Nichrome wire, what is the current in the wire?
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Kirchhoff’s Rules
1. Junction rule. The sum of the currents entering any junction in a
circuit must equal the sum of the currents leaving that junction:
2.Loop rule. The sum of the potential differences across all
elements around any closed circuit loop must be zero:
Series and Parallel Resistors
series combination
For a series combination of two resistors, the currents are the same in
both resistors because the amount of charge that passes through R 1
must also pass through R2 in the same time interval.
V
 R1  R2
I
Req  R1  R2
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series combination
The equivalent resistance of three or more resistors
connected in series is
This relationship indicates that the equivalent resistance of
a series connection of resistors is the numerical sum of the
individual resistances and is always greater than any
individual resistance.
Example:
With the switch in the circuit of Figure closed (left), there is no current in R2, because the current
has an alternate zero-resistance path through the switch. There is current in R1 and this current is
measured with the ammeter (a device for measuring current) at the right side of the circuit. If the
switch is opened (Fig, right), there is current in R2 . What happens to the reading on the ammeter
when the switch is opened? (a) the reading goes up; (b) the reading goes down; (c) the reading
does not change.
(b) the reading goes down
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A parallel connection
When resistors are connected in parallel, the potential differences
across the resistors is the same.
I
1
1
(

)
V
R1 R2
1
1
1
(  )
Req
R1 R2
A parallel connection
An extension of this analysis to three or more resistors in parallel
gives:
the inverse of the equivalent resistance of two or more resistors connected
in parallel is equal to the sum of the inverses of the individual resistances.
Furthermore, the equivalent resistance is always less than the smallest
resistance in the group.
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Example:
With the switch in the circuit of Figure open (left), there is no current in
R2. There is current in R1 and this current is measured with the ammeter
at the right side of the circuit. If the switch is closed (Fig., right), there is
current in R2. What happens to the reading on the ammeter when the
switch is closed?
(a) the reading goes up; (b) the reading goes down; (c) the reading does
not change.
(a) the reading goes up
Example Find the Equivalent Resistance
Four resistors are connected as shown in Figure.
(A) Find the equivalent resistance between
points a and c.
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(B) What is the current in each resistor
if a potential difference of 42 V is
maintained between a and c?
Example:
Consider the circuit shown in Figure. Find (a) the current in the
20.0Ω resistor and (b) the potential difference between points a
and b.
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Kirchhoff’s Rules in complex circuits
•Because charges move from the high-potential end of a resistor
toward the lowpotential end, if a resistor is traversed in the direction
of the current, the potential difference ΔV across the resistor is -IR
(Fig. a).
• If a resistor is traversed in the direction opposite the current, the
potential difference Δ V across the resistor is +IR (Fig. b).
Kirchhoff’s Rules in complex circuits
• If a source of emf (assumed to have zero internal resistance) is
traversed in the direction of the emf (from - to +), the potential
difference Δ V is + (Fig. c). The emf of the battery increases the
electric potential as we move through it in this direction.
• If a source of emf (assumed to have zero internal resistance) is
traversed in the direction opposite the emf (from + to - ), the
potential difference Δ V is - (Fig. d). In this case the emf of the
battery reduces the electric potential as we move through it.
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Example Applying Kirchhoff’s Rules
Find the currents I1, I2, and I3 in the circuit shown
in Figure
Applying Kirchhoff’s junction rule to junction c gives
We now have one equation with three unknowns—I1, I2, and I3. There are three
loops in the circuit—abcda, befcb, and aefda. We therefore need only two loop
equations to determine the unknown currents. (The third loop equation would
give no new information.) Applying Kirchhoff’s loop rule to loops abcda and
befcb and traversing these loops clockwise, we obtain the expressions
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Note that in loop befcb we obtain a positive value when traversing the 6.0 Ωresistor
because our direction of travel is opposite the assumed direction of I1.
Expressions (1), (2), and (3) represent three independent equations with three
unknowns. Substituting Equation (1) into Equation (2) gives
Dividing each term in Equation (3) by 2 and rearranging gives
Subtracting Equation (5) from Equation (4) eliminates I2, giving
Using this value of I1 in Equation (5) gives a value for I2:
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