Physics 122B
Electricity and Magnetism
Lecture 16 (Knight: 31.9,10 & 32.1)
Grounding, RC Circuits, Magnetism
May 7, 2007
Martin Savage
Lecture 17 Announcements
 Lecture HW Assignment #5 has been
posted on Tycho and is due at 10 PM on
Wednesday.
 Check Tycho for your exam scores. If
there are missing parts, you may not have
put your name on your paper. See Helen
Gribble, C136 PAB, to fix such problems.
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Physics 122B - Lecture 17
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Example:
Analyzing a Complex Circuit (2)
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Maximum Power Transfer
Question: For a battery with EMF E and
internal resistance r, what value should an
adjustable external load resistance R have
to make the power dissipated in R as large
as possible?
E
I
rR
I
r
2
2
E
E
R


PR  I 2 R  
R


2
r

R


r  R
dPR
E
E R
rR
2


2

E
0
2
3
3
dR  r  R 
r  R
r  R
2
2
R
E
PR is a maximum when r-R = 0 or r=R. In other words, energy
can be drawn from the battery at the greatest rate when the external
load resistance matches the internal resistance of the battery. This is
called load matching. It is very important in transferring energy and
signals with minimum loss. But note that only ½ of the energy gets to R.
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Grounding and GFI
Modern power wiring includes a “ground” line, the
round 3rd wire of an electrical plug. The ground point
defines a point of zero potential, which is normally
connected directly to the Earth (Vearth=0). The
operation of any circuit depends only on potential
differences, so it should not be affected by the
presence or absence of a ground connection.
Because the ground connection is connected at
only one point, no current should flow through the
ground connection. However, if some other part of a
circuit is accidentally grounded, current is likely to
flow through the ground line.
GFI (ground fault interruption) circuits, widely
used, e.g., in bathroom wiring, detect current flow in
the ground line and interrupt power automatically
when it occurs. This has prevented many accidental
electrocutions.
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Light Fixture
V=0
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Example:
A Grounded Circuit
The circuit shown is grounded at
the junction between the two
resistors rather than at the bottom.
Find the potential at each corner
of the circuit.
E
10 V
I 
 0.5 A
R 8   12 
V1  (8 )(0.5 A)  4 V
V2  (12 )(0.5 A)  6 V
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RC Circuits
I = - dQ/dt
Q
Q
dQ
VC  VR   IR   R
0
C
C
dt
Qf
t
ln

Qi
RC
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dQ
1

dt
Q
RC
 t 
 exp  
Qi
 RC 
Qf
Physics 122B - Lecture 17
Qf

Qi
t
dQ
1

dt

Q
RC 0
Q f  Qi et / RC
Exponential decay!
8
RC Exponential Decay
Q(t )  Q0e
 t / RC
 Q0e
t /
Define RC time constant:   RC
1/ e  1/ 2.71828  0.367879

t
RC
dQ(t ) Q0
I (t ) 

e
dt
RC
 I 0 et / RC  I 0et /
I0 Q0/RC
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Charging Capacitors:
Early and Late
Initially, when a switch closes there is a potential difference of 0 V
across an uncharged capacitor. After a long time, the capacitor reaches
its maximum charge and there is no current flow through the capacitor.
Therefore, at t=0 the capacitor behaves like a short circuit (R=0), and at
t=∞ the capacitor behaves like at open circuit (R=∞).
Example:
100 V
100 V
12.5 A
100 V
2.5 A
Circuit
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10 A
0 V
at t=0
at t=∞
Calculate initial currents.
IB = 100 V/8  = 12.5 A
Calculate final potentials.
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Example: Exponential
Decay in a RC Circuit
The switch has been in position
``a’’ for a long time. It is changed to
position ``b’’ at t=0.
What are the charge is the
capacitor and the current through the
resistor at t=5.0 ms?
  RC  (10 )(1.0 10-6 F)  10 ms
Q0  C VC  (1.0 10-6 F)(9.0 V)  9.0 10-6 C
Q(5 ms)  Q0 e  t / RC  (9.0 10-6 C) exp(.5)  5.5 m C
Q0 (9.0 10-6 C)
I0 

 0.90 A
RC
(10 ms)
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I (5 ms)  I 0e t / RC  (0.90 A) exp( .5)  0.55 A
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Charging a Capacitor
Using the Loop Law: E  R
Try: Q  a  be

t
RC
dQ 1
 Q0
dt C
dQ
b  RCt
, so

e
dt
RC
b  RCt
a b  RCt
E  e
  e
 a EC
C
C C
At t  0, I 
dQ E
b
 
 b  E C
dt R
RC
t
t






Therefore, Q  E C 1  e RC   Qmax 1  e RC 




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Question
The time constant for the discharge of the capacitor is:
(a) 5 s;
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(b) 4 s;
(c) 2 s;
(d) 1 s; (e) the capacitor does not discharge
because the resistors cancel.
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Plumber’s RC Analogy*
Valve
Constriction
P1
Pump
The “plumber’s analogy” of an RC circuit is a
pump (=battery) pumping water in a closed
loop of pipe that includes a valve (=switch), a
constriction (=resistor), and a rubber
diaphragm. When the valve starts the flow,
the diaphragm stretches until the pressure
difference across the pump (P1-P3) equals
that across the diaphragm (P2-P3).
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P2
Rubber
Diaphragm
P3
Pump = Battery
Valve = Switch
Constriction = Resistor
Capacitor= Rubber Diaphragm
Pressure = Potential
Water Flow = Current
14
Chapter 31 Summary (1)
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Chapter 31 Summary (2)
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Chapter 31 Summary (3)
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Experiments with Magnetism:
Experiment 1
Tape a bar magnet to a cork
and allow it to float in a dish of
water.
The magnet turns and aligns
itself with the north-south
direction.
The end of the magnet that
points north is called the
magnet’s north-seeking pole, or
simply its north pole. The
other end is the south pole.
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Experiments with Magnetism:
Experiment 2
Bring the north poles of two bar magnets near to each other. Then
bring the north pole of one bar magnet near the south pole of another
bar magnet.
When the two north poles are brought near, a repulsive force
between them is observed. When the a north and a south pole are
brought near, an attractive force between them is observed.
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Experiments with Magnetism:
Experiment 3
Bring the north pole of a bar magnet near a compass needle.
When the north pole is brought near, the north-seeking pole of the
compass needle points away from the magnet’s north pole. Apparently
the compass needle is itself a little bar magnet.
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Experiments with Magnetism:
Experiment 4
Use a hacksaw to cut a bar magnet in half. Can you isolate the north
pole and the south pole on separate pieces?
No. When the bar is cut in half two new (but weaker) bar magnets
are formed, each with a north pole and a south pole. The same result
would be found, even if the magnet was sub-divided down to the
microscopic level.
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Experiments with Magnetism:
Experiment 5
Bring a bar magnet near an assortment of objects.
Some of the objects, e.g. paper clips, will be attracted to
the magnet. Other objects, e.g., glass beads, aluminum foil,
copper tacks, will be unaffected. The objects that are
attracted to the magnet are equally attracted by the north
and south poles of the bar magnet
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Experiments with Magnetism:
Experiment 6
Bring a magnet near the
electrode of an electroscope.
There is no observed effect,
whether the electroscope is
charged or discharged and
whether the north or the south
pole of the magnet is used.
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Conclusions from Experiments
1.
2.
3.
4.
5.
Magnetism is not the same as electricity. Magnetic poles
are similar to charges but have important differences.
Magnetism is a long range force. The compass needle
responds to the bar magnet from some distance away.
Magnets have two poles, “north” (N) and “south” (S). Like
poles repel and opposite poles attract.
Poles of a magnet can be identified with a compass. A
north magnet pole (N) attracts the south-seeking end of
the compass needle (which is a south pole).
Some materials (e.g., iron) stick to magnets and others do
not. The materials that are attracted are called magnetic
materials. Magnetic materials are attracted by either
pole of a magnet. This is similar in some ways to the
attraction of neutral objects by an electrically charged
rod by induced polarization.
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Monopoles and Dipoles
Every magnet that has ever
been observed is a magnetic
dipole, containing separated
north and south poles. Attempts
to isolate one pole from the
other fail.
It is theoretically possible to have magnetic monopoles, i.e., isolated
magnetic poles with a “north” or “south” magnetic charge. Search have
been conducted, but no such object has ever been found in nature. For
the purposes of this course, we will assume that isolated magnetic
monopoles do not exist, but we will point out the places in the formalism
where they would go if they did exist.
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Lecture 17 Announcements
 Lecture HW Assignment #5 has been
posted on Tycho and is due at 10 PM on
Wednesday.
 Check Tycho for your exam scores. If
there are missing parts, you may not have
put your name on your paper. See Helen
Gribble, C136 PAB, to fix such problems.
7/19/2016
Physics 122B - Lecture 17
26