Transcript Chapter 31

Reading Quiz – Chapter 31
1. Name or describe Kirchhoff’s Laws.
2. Which of the following are ohmic
materials?
a.
b.
c.
d.
e.
Batteries
Wires
Resistors
Light bulb
capacitors
f. Materials a-d
g. Materials b and c
h. Materials a and b
i. All of a-e
Learning Objectives – Ch 31
• To develop a conceptual model of simple
circuits.
• To understand series and parallel
resistances.
• To apply Kirchhoff’s laws to circuit analysis.
• To find the connection between current and
potential difference for a conductor.
• To understand energy transfer and power
dissipation in circuits.
• To understand RC circuits.
ε and ∆V
• For an ideal battery
potential difference
between the positive and
negative terminals, ∆V,
equals the chemical work
done per unit charge.
∆V = Wch /q = ε
• ε is the emf of the battery
• Due to the internal
resistance of a real
battery, ∆V is often
slightly less than the emf.
Ch – 31 Circuits
Resistors and Ohm’s Law
From the relationship
J = σE = E/ρ
where ρ, resistivity = 1/ σ
It can be shown that:
I = ∆V/R
where R, resistance = pL/A
This is Ohm’s Law
Units of resistance are Ω
(ohms)
Ohmic and nonohmic materials
Ohm’s Law is only valid
for ohmic materials –
where the current is
proportional to the
potential difference
Question
• Calculate the resistance of a 1.0 mm-diameter,
3 cm-long carbon resistor. Table 30.1, p.944
gives resistivity values:
Answer: 1.34 Ω
• Calculate the resistance of a 1.0 mm-diameter,
3 cm-long carbon resistor. Table 30.1, p.944
gives resistivity values:
Ohmic materials: the “Ideal” wire
• Wires are metals with
a very very small
resistance (R<< 1 Ω)
• The ideal wire has a
R=0 Ω, therefore
∆V = 0 V between the
ends of an ideal wire.
Hmmm….
•
Recall that Ohm’s
Law is I = ∆V/R
• If the wire at the
right is an ideal wire,
will the value of the
current be:
a. undefined and
therefore 0
b. Huge
Ohmic materials: Resistors
• Resistors are conductors, albeit poor ones, that
are used so the current in a circuit of ideal wires
will not be “huge”. Typical Range of R: 10 to
106 Ω.
• A light bulb filament is a non-ohmic resistor.
Ohmic Materials - Insulators
•
•
Insulators are materials such as glass, plastic
or the space between two parts of a wire after
you have removed a resistor or light bulb (open
circuit). The ideal insulator has a resistance of
infinity.
Recall that Ohm’s Law is I = ∆V/R . Will an
open circuit result in:
1. current of zero
2. potential difference of zero
3. both values going to zero
Kirchoff’s Loop Law
• The sum of all
potential differences
encountered while
moving around closed
path is zero
• This is a result of the
conservation of
energy for a
conservative force
Kirchoff’s Junction Law
• The total current
into a junction must
equal the total
current out of the
junction
• Based on
conservation of
charge and current
Σ Iin = ΣIout
Basic Circuit for the Carbon
Resisitor
• Calculate the current
that leaves the
positive terminal of
the ideal battery (ε=
4V).
• Is this current greater
than, less than or
equal to the current
at leaving the bottom
end of the load?
Answers
• 2.99 A
• Equal to: Current out
of the battery must
equal current into it
according to Kirchoff’s
Junction Law;
Energy and Power
Power is that rate at which energy is
transferred
Pbattery = rate at which the battery supplies
energy to the charges
Pbattery = of energy transfer = dU/dt
Pbattery = (dq/dt)ε
Pbat = I ε
What happens to the energy?
• The chemical energy of the
battery is converted to U,
electrical potential energy:
Echem  U
• The resulting electric field
causes the electrons to
accelerate: UK
• Collisions in the lattice
structure transfer the energy to
the lattice as thermal energy:
KEth
• Thermal energy is a
dissapative energy (i.e. can’t
be recovered like mechanical
energy.
What happens to the energy?
• The rate at which
energy is transferred
from the current to
the resistor is:
PR = dEth/dt
PR = (dq/dt)∆VR
PR = I ∆VR
• Since a resistor
obeys Ohm’s Law:
PR = I2R = (∆VR)2/R
Numerical question
What is the magnitude
and direction is the
current in the 30 Ω
resistor?
Draw a graph of the
potential as a function
of the “distance”
traveled through the
circuit (see ex 31.2 for
distance
Answer
a. Current = 0.100 A
Power rating of a light bulb or
household applicance
• Commercial and residential electrical systems
are set up so that each individual appliance
operates at a potential difference of 120 V.
• Power Rating or Wattage is the power that the
appliance will dissipate at a potential difference
of 120 V.
• Power consumption will differ if operated at any
other potential difference (i.e. 220, such as is
standard in Europe and many other countries).
Resistors in Series
• Resistors aligned from end to end, with no junctions between them
are in series
• The current I is the same through each of the resistor
• The potential difference through each resistor is :
∆VR = (-)IR
The equivalent resistor is made by adding up all resistors in series in
the circuit.
Resistors in series question
• What is the Value of R?
Answer: R = 25Ω
Real Batteries
• Real batteries have
an internal resistance,
r, and therefore:
∆V = (ε – Ir)< ε
∆V = R/( R +r) ε
#19 in workbook
Workbook Problem #14
A 60 W light bulb and a 100
W bulb are placed one
after the other in a circuit.
The battery’s emf is 120V.
Which one glows more
brightly? Assume the
resistance of the bulbs to
be ohmic.
Answer
• The 60 W bulb.
• Calculate the resistance of each bulb,
knowing the power rating when it is
operated alone at 120V. Which has the
higher resistance?
• Calculate the equivalent resistance, and
the current they both draw when in the
series circuit
Resistors in Parallel
• Resistors that are connected at
both ends are considered to be
in parallel.
• Connecting resistors in this
manner increases the amount
of current since it increases the
number of paths.
• The left end of R1 has the
same potential as the left end
of R2 and the same is true of
the right end. Therefore the
potential difference (∆V)
across each resistor connected
in parallel is equal.
Resistors in Parallel
• Kirchoff’s current law
applies at the
junctions. Ic = Id.
Therefore the current
through each of the
branches depends on
the resistance of the
resistor
• The equivalent
resistance:
Rcd = (1/R1 + 1/R2)-1
Parallel resistor question
What is the value of R?
Answer
• IR = .667A
• R = 12 Ω
Complex resistor problem
• Find the current through the circuit.
• What are the values of potential at points a-f?
Complex resistor problem
• Answer: 1A
• a:0V b:12V c:9V d: 6V e: 0V f: 0V
The circuit is grounded at point c
What is the current through the circuit?
What are the values of potential at points a-f?
The circuit is grounded at point c
What is the current through the circuit? – still 1A
What are the values of potential at points a-f?
a: -9V b: 3V c: 0V d: -3V e: - 9V, f: -9V
The Capacitance of a Capacitor
(Ch. 30)
• 2 expressions for the electric field of a
capacitor:
E = Q/(ε0 A)
relates electric field to surface charge
density
E = ∆Vc/d
Electric field is the rate of change of the
potential across plates
The Capacitance of a Capacitor
(Ch. 30)
• Combining the 2 expressions:
Q = [(ε0 A)/d]∆Vc
• potential difference of the capacitor is
proportional to the charge on the plates.
• The constant of proportionality is a
geometric factor
The Capacitance of a Capacitor
(Ch. 30)
• Q = [(ε0 A)/d]∆Vc
• Capacitance (C) = (ε0 A)/d. For a given
charge, a capacitor with a larger
capacitance will have a greater potential
difference.
• The SI unit of capacitance is the farad:
1 farad = 1 F = 1 Coulomb/Volt.
• Typical capacitors have values in the
microfarad to picofarad range.
RC Circuit
• Consists of a capacitor
and a resistor.
• Unlike a battery, a
capacitor cannot provide
a constant source of
potential difference.
• This value is constantly
changing as the charge
leaves the plate.
• Current due to a
discharging capacitor is
finite and changes over
time.
RC Circuit
•
Use Kirchoff’s loop law to
analyze the circuit, starting at
the capacitor:
∆Vc + ∆VR = 0
Q/C – IR = 0
•
•
•
Since charge is being
removed from the plates:
I = - dQ/dt
Q/RC + dQ/dt =0
This is a differential equation,
but is relatively easy to solve
by rearranging terms:
dQ/Q = -[1/(RC)]dt
And then by integrating….
RC Circuit
• The initial conditions
were Q = Q0 at t=0
(switch closes)
Q
 dQ / Q
Qo
t
= -1/RC
 dt
0
ln Q – ln Q0 = -t/RC
ln (Q/Q0) = - t/RC
RC Circuit
ln (Q/Q0) = - t/RC
To solve for Q, capacitor
charge at time t, take the
exponential of both
sides and multiply by Q0:
Q = Q0e-t/RC
the quantity RC is called
the time constant, τ:
Q = Q0e-t/ τ
I = dQ/dt = I0e-t/τ
Charging a capacitor
• After the switch is closed, the
battery will move charge from
the bottom plate to the top plate
of the capacitor.
• As the charge on the capacitor
increases, the current through
the circuit will decrease.
• Current ceases when ∆Vc = ε.
Q = Qmax(1 – e-t/τ)
Capacitor Problem
• A 10 μF capacitor initially charged to 20
μC is discharged through a 1 kΩ resistor.
How long does it take to reduce the
capacitor’s charge to 10 μC?
Capacitor Problem
• A 10 μF capacitor initially charged to 20
μC is discharged through a 1 kΩ resistor.
How long does it take to reduce the
capacitor’s charge to 10 μC?
Answer: 10 μC = (20 μC) et/0.010 s
Take the natural logarithm of both sides:
ln (10 μC/20 μC) = - t/0.010 s
= 6.93 s