Study Guide 8: Electricity
Text: chapter 15, sections 2-3 and 5.
(Note, the order in which we cover this material
differs from the Study Guide, see Course Outline.
We start with section 15.5)
Quiz 5 (final day, Friday, March 28, 2008) you
should know:
Material from Computer lab exp. 7 (Radiation) and exp. 8
(Electricity): in particular
From semilog and log-log graphs of provided data (exp.
7);
1. How to calculate the linear attenuation coefficient.
2. Determine half-time T1/2.
3. Study the exponent in power-law decay from a point
radioactive source.
4. Apparatus used in exp. 8 (bottom of page 8-1 of lab
manual) and how to arrange this to study parallel and
series circuits.
5. Using 4 to calculate current and resistances.
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ELECTRIC CURRENT
The 3 quantities describing a DC circuit are:
1) Current (I)
2) Electric Potential (V)
3) Resistance(R)
Definitions:
Current=flow of charge, Δq, through a cross-sectional
area in time Δt.
I = Δq/Δt
SI unit = Ampere (A) = (1 Coulomb/second).
Motion of electrons (negative!) actually produces the
current. E.g., in a copper wire:
2
Copper atomic cores (ignore their motion)
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 in moving a charge q from a point B to a
point A is WAB, then
VAB = WAB/q
where VAB = VA−VB
SI unit = Joules/Coulomb = Volt (V). Electric potential is
often called the voltage.
3
Resistance = applied voltage required per current depends on material.
Consider a voltage, VAB, across a material with resistance
R:
R
A
B
I
VAB = VA − VB
If R is a constant, the above quantities obey
VAB = IR .
UNITS FOR R:
Ohm's Law
Volt/Amperes = Ohms (Ω).
(Conductance (unit: Siemens, S = Ω−1) is the inverse of resistance,
g = 1/R
)
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DC CIRCUITS:
• Current varies sinusoidally on power grids - Alternating
Current (AC)
• We study constant current - Direct Current (DC).
R
A simple DC circuit
I
+
−
V
(note the + terminal of battery is
at higher potential than the −)
Kirchoff's rules
1.
The current entering a point is equal to the current
leaving it.
2.
The net change in voltage around a closed loop is
zero.
Other useful observations:
• points connected by a PERFECT resistor are at the
same electric potential.
5
• Resistors arranged in parallel and series have specific
rules to determine their effective resistance.
PARALLEL NETWORK:
R1
I1
R2
I
I
A
I2
R3
I3
V
The equivalent resistance, Req, such that V = IReq, for the
3 resistors above is found using Kirchoff's first rule at
point A,
Current in
I = I 1 + I2 + I 3
Current out
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NB All 3 resistors have same voltage across them, V,
(parallel). Therefore
V = I1R1
V = I2R2
V = I3R3
∴ I = V/R1 + V/R2 + V/R3
= (1/R1 + 1/R2 + 1/R3)V
Req = (1/R1 + 1/R2 + 1/R3)-1
so that
SERIES NETWORK:
I
R1
R2
R3
A
B
V1
V2
V3
VAB = VA−VB
Ohm’s law: IReq = VAB. Notice…
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1. Each resistor now experiences a different V.
2. VAB = V1+V2+V3
3. Current out = current in, so each resistor has the
same current I.
Ohm’s law for each resistor;
IR1 = V1
IR2 = V2
IR3 = V3
∴ IReq = IR1 + IR2 + IR3 .
Dividing by I on both sides, we find the equivalent
resistance,
Req = R1 + R2 + R3
8
Sample Problem 1: Parallel Resistors
Calculate Req for the following network of resistors:
R1 = 50.0 Ω
R2 = 33.3 Ω
R3 = 16.6 Ω
As shown earlier, Req = (1/R1 + 1/R2 + 1/R3)-1. Therefore,
Req = (1/50.0 + 1/33.3 + 1/16.6)−1 = (0.02 + 0.03 + 0.06)−1.
That is, Req = 9.1 Ω.
Sample Problem 2: Resistors in Series
Calculate Req for the same three resistors in a series
arrangement:
R2
R3
R1
Ans. Req = 99.9 Ω.
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Sample Problem 3: Series/Parallel Resistors
Calculate Req for the following combination of
resistors:
R1 = 50.0 Ω
R3 = 100.0 Ω
R2 = 25.0 Ω
R4 = 166.7 Ω
Ans. Req = 79.2 Ω.
Sample Problem 4: Series, Parallel or
What?
Calculate Req for the following network. All the resistors
are 2.0 Ω, while the wires are assumed to be perfect
resistors.
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This is a parallel-series arrangement. Using methods
similar to example 3 to solve for Req, we find that Req =
5/3 = 1.67 Ω.
CALCULATING THE RESISTANCE:
Resistance depends on the dimensions and material
through which the current is flowing. For example,
consider a conductor with cross-sectional area A and
length L;
L
A
Then, the resistance is given by,
R=
ρL
A
where ρ is the resistivity of the material and can be
found in tables of scientific data. For metals ρ ≈ 10-8
Ω⋅m; for insulators ρ ≈ 10+13 Ω⋅m (more than 10+20 times
that of a metal!).
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As hinted earlier, resistivity depends upon the collisions
of the current carriers with the ionic cores making up
the bulk of the material. Clearly this will depend on the
temperature of the material. The higher the
temperature, the greater the ionic core oscillations.
These oscillations impede the flow of electrons
contributing to the resistance, so usually the resistivity
increases with temperature.
Sample Problem 5: Resistance of a cylinder
Calculate the resistance of the cylinder below
(be careful about the units !):
2.00 cm
(diameter) 0.500 mm
ρ = 1.50 × 10-2 Ω⋅cm
Ans. R = 15.3 Ω
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For additional problems involving circuits
with batteries, recall Kirchoff’s rules:
1.
The current entering a point is equal to the current
leaving it.
(The “point rule”)
2.
The net change in voltage around a closed loop is
zero.
(The “loop rule”)
These rules must be supplemented by important sign
conventions (see text page 15-19).
Example
3V
1Ω
20 Ω
e
f
loop 1
I2
I2
I2
c
d
5V
I3
1Ω
I1
loop 2
I1
a
I3
b
5Ω
I3
(a) What is the current, I1, through the middle
branch?
(b) What is Vab = (Va - Vb)?
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GENERAL STRATEGY – MESH EQUATIONS: Use
Kirchoff's rules and Ohm's Law to solve for the unknown
loop currents, I1, I2, and I3, then determine Vab.
(1)
Use Kirchoff's Point Rule (#1) to relate currents.
E.g., at point c:
I1 = I2 + I3
or I3 = I1 - I2
(2) Use Kirchoff's Loop Rule (#2) on as many loops
as required to generate sufficient equations. Two
loops are required from above.
(i) For loop 1: going clockwise starting at c (but
this choice of direction is arbitrary)
-3V - I2 (1 Ω) - I2 (20 Ω) - I1 (1 Ω) + 5V = 0
“-” because going from + to – terminal of the
battery
The IR terms are negative since each
resistor is traversed in the same direction as
the current (this is one of the sign
conventions).
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∴ - I1 - 21 I2 = -2 Amps
∴ I1 + 21 I2 = 2 A
(ii)
[1]
For loop 2: going clockwise starting at c (and
using I3 = I1−I2)
- 5 V + I1(1 Ω) + (I1 - I2)(5 Ω)
= 0
Now the IR terms are positive because
each resistor is traversed in the
opposite direction to the current.
∴ 6 I 1 - 5 I2 = 5 A
[2]
(3) Solve the resulting system of equations:
Multiplying equation [1] by 5:
5 I1 + (5)(21)I2 = 10 A
[3]
Multiplying equation [2] by 21:
(21)(6) I1 − (5)(21)I2 = 105 A
[4]
Add [3] and [4] (terms in I2 cancel) …
131 I1= 115 A
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∴ I1 = 115/131 = 0.878 A
(ans. To (a))
We’ll also need I2: from [2], I2 = (6I1 − 5)/5
= (6)0.878/5 -1 = 0.0536 A
For part (b), Vab = (I1 − I2) (5 Ω) = 4.12 V. This is also
equal to Vcd and Vef because these voltages are all
parallel.
(Check: Vcd = 5V − I1(1 Ω ) = 5V − 0.878V = 4.12 V
(The IR term is negative because I1 is toward the left.)
Vef = 3V + I2(1Ω +20Ω) = 3V + 0.0536(21) = 4.126 V)
ELECTRIC FORCE AND FIELD
The underlying concepts from the last section are
somewhat explained. E.g., Ohm's law stems from the
electric forces between two charges (Coulomb's law)
General Information:
• The SI unit of charge is the Coulomb (C).
• The smallest magnitude of charge that can occur
outside of a nucleon is that on a single electron, =
1.602 × 10 -19 C.
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• Charge is either positive or negative. Charges of the
same sign (-,- or +,+) repel and charges of opposite sign
(+,- or -,+) attract.
ELECTRIC FORCE-COULOMB'S LAW:
q1
F21
F12
q2
NB: Equal attracting force on EACH charge! (These
forces are in opposite directions due to Newton’s 3rd law)
is given by the signs of the charges.
2
F1
• direction of
G
• magnitude is given by Coulomb's law.
• If there are more charges, add forces together as
VECTORS. (But, although there are examples in the
text and study guide where the charges are
arranged in a non-linear manner (e.g., example 15-1 in
text and on page SG 8-5), in practice we only deal
with linear arrangements of charges, which involves
only algebraic addition of forces.
Assuming nothing (not even air) lies between the charges:
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G
k q1q2
F12 =
2
(r12 )
where k = 9.0 × 109 Nm2C-2 is the constant for the
vacuum. If the charges are not in a vacuum then
Coulomb's law takes the form:
G
k q1q2
F12 =
2
K ( r12 )
where K is the (unitless) dielectric constant for the
medium in which the charges are immersed. For example:
water
air
wood
K = 80.4
K = 1.0005
K = 5.0
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ELECTRIC FIELDS:
Each charge creates an electric field around itself. The
interaction of a second charge with this electric field
gives rise to the electric force. For example,
Electric field lines:
The magnitude of the electric field at distance r from a
point charge q is given by:
G kq
E=
2
Kr
where k and K have the same meanings as in Coulomb’s
Law. The direction of the field = direction of the force
on small positive charge at that point.
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If there are multiple sources, the total field = the
VECTORIAL sum of the electric fields from all sources.
FORCE ON A CHARGE q1 IN AN ELECTRIC FIELD E:
G
G
F = q1E
.
NB: electric field and force point in the same direction
for a positive charge q1 and in the opposite direction for
a negative charge.
ELECTRIC POTENTIAL ENERGY and
POTENTIAL:
Consider a small positive charge q1 in a constant electric
field E. It will experience a force, q1E, in the direction
of E. To move it in the opposite direction, a force with
magnitude q1E must be applied in that direction. To move
the charge a distance d in that direction, work must be
done against the electric force. Work(W) = force ×
distance for constant forces.
∴ W=Fd=q1Ed
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This energy depends on the charge q1. The electric
potential, V, is defined to be independent of that
charge, i.e.,
V = W/q1
or
W = q1V ( = electric potential energy U)
Potential due to a point charge
The field due to a point charge isn't constant, and to get
the work required to place two charges a distance r apart
you must integrate the force over the path traveled.
A point of reference must be chosen where V = 0. For
point charges r = ∞ is assumed to have V = 0. The
electric potential around a point charge q surrounded by
a medium with dielectric constant K then takes the form,
V
=
kq
Kr
The energy U (or work) required to move q1 from infinity
to a distance r from q is
U = q1V
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Using the previous equation for the potential around q,
we get
U =
k qq
1
K r
.
Finally, the potential at a position due to many point
charges qi is the algebraic sum of the potentials from
each charge, i.e.,
V=
∑V
i
i
Sample Problem 6
A sodium ion (singly charged positive ion) is 75 nm
(1nm = 10-9m) from a chloride ion (singly charged
negative ion) in the protoplasm of a cell (assume the
protoplasm has the same dielectric constant as water).
What forces do the ions exert on each other?
Note that the sodium ion has a positive charge, qNa+ =
1.602 × 10 –19 C and the chloride ion has a negative
charge, QCl− = -1.602 × 10 –19 C.
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Ans. F = 5.11 × 10−16 N
This force is attractive (since the charges are of
opposite sign), hence the forces on each particle have
the directions shown on page 17.
Sample Problem 7
Three ions (Ca2+, Na+ and Cl−) in a vacuum (hence K =
1) are in a linear arrangement shown below:
Ca2+
Na+
r12 =20 nm
Cl−
r23 =10 nm
Find
(a) The net electric field E at the position of the Na+
ion.
(b) The net force on the Na+ ion.
(c) The electric potential V at the position of the Na+
ion.
(d) Repeat (a)−(c) at the position of the Cl− ion.
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Answers:
(a) E = 2.16×107 N/C (acting toward the right)
(b) F = 3.46×10−12 N (acting toward the right)
(c) V = 0.
(d) (Without numbers)
E= ECa2+ + ENa+ =
kqCa2+
(r
12
+r23 )
2
+
kqNa+
2
r23
(toward right)
F = qCl−E (toward left)
V=
kqCa2+
(r
12
+r23 )
+
kqNa+
r23
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