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my office!
Announcements
 Special Homework #6 is due tomorrow. You can download it
here, if necessary:
http://campus.mst.edu/physics/courses/24/Handouts/special_homework.html
Today is our last day on circuits. By the
end of today, you should have mastered
the elementary aspects of dc circuits,
including debugging and fixing broken
ones.
Announcements
Exam 2 is two weeks from yesterday.
Contact me by the end of next Wednesdays lecture if you
have special circumstances different than for exam 1.
A note on homework…
Today’s agenda:
Measuring Instruments: ammeter, voltmeter,
ohmmeter.
You must be able to calculate currents and voltages in circuits that contain “real”
measuring instruments.
RC Circuits.
You must be able to calculate currents and voltages in circuits containing both a resistor
and a capacitor. You must be able to calculate the time constant of an RC circuit, or use
the time constant in other calculations.
Measuring Instruments: Ammeter
You know how to calculate the
current in this circuit:
I=
R
V
.
R
r
If you don’t know V or R, you can
measure I with an ammeter.
V
Any ammeter has a resistance r. The current you measure is
V
I=
.
R +r
To minimize error the ammeter resistance r should very small.
Example: an ammeter of resistance 10 m is used to measure
the current through a 10  resistor in series with a 3 V battery
that has an internal resistance of 0.5 . What is the percent
error caused by the nonzero resistance of the ammeter?
R=10 
Actual current:
I=
V
R +r
3
I=
10 + 0.5
r=0.5 
V=3 V
You might see the symbol 
used instead of V.
I = 0.286 A = 286 mA
Current with ammeter:
V
I=
R +r +R A
I=
3
10 + 0.5+ 0.01
I = 0.285 A = 285 mA
% Error =
0.286 - 0.285
 100
0.286
% Error = 0.3 %
R=10 
r=0.5 
RA
V=3 V
Not bad in a Physics 24 lab!
A Galvanometer
When a current is passed through a coil connected to a needle,
the coil experiences a torque and deflects. See the link below
for more details.
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html#c1
An ammeter (and a voltmeter—coming soon) is based on a
galvanometer. OK, everything is electronic these days, but the principles here still apply.
We’ll learn about galvanometers later.
For now, all you need to know is that the
deflection of the galvanometer needle is
proportional to the current in the coil (red).
A typical galvanometer has a resistance of a few tens of ohms.
Hold it right there. Didn’t you say an ammeter must have a very
small resistance. Is there a physics mistake in there
somewhere?
A galvanometer-based ammeter uses a galvanometer and a
shunt, connected in parallel:
RG
G
IG

A
RSHUNT
I
I
ISHUNT
Everything inside the blue box is the ammeter.
The resistance of the ammeter is
1
1
1


R A R G R SHUNT
R G R SHUNT
RA 
R G  R SHUNT
RG
I
A
IG
G
RSHUNT
B
ISHUNT
Homework hint: “the galvanometer reads 1A full scale”
means a current of IG=1A produces a full-scale deflection of
the galvanometer needle. The needle deflection is
proportional to the current IG.
If you want the ammeter shown to read 5A full scale, then
the selected RSHUNT must result in IG=1A when I=5A. In that
case, what are ISHUNT and VAB (=VSHUNT)?
AExample:
galvanometer-based
ammeter uses
a galvanometer
and a to
what shunt resistance
is required
for an ammeter
shunt,
have aconnected
resistanceinofparallel:
10 m, if the galvanometer resistance is
60 ?
RG
1
1
1


R A RG RS
1
1
1


RS R A RG
I
R G R A  60  .01
RS 

 0.010 
R G -R A
60 -.01
(actually 0.010002 )
IG
G
RS
IS
The shunt resistance is chosen so that IG does not exceed the
maximum current for the galvanometer and so that the
effective resistance of the ammeter is very small.
R G R A  60  .01
RS 

 0.010 
R G -R A
60 -.01
To achieve such a small resistance, the shunt is probably a
large-diameter wire or solid piece of metal.
Web links: ammeter design, ammeter impact on circuit,
clamp-on ammeter (based on principles we will soon be
studying).
Measuring Instruments: Voltmeter
You can measure a voltage by placing a galvanometer in
parallel with the circuit component across which you wish to
measure the potential difference.
RG
G
Vab=?
a
R=10 
r=0.5 
V=3 V
b
Example: an galvanometer of resistance 60  is used to
measure the voltage drop across a 10 k resistor in series with
a 6 V battery and a 5 k resistor (neglect the internal
resistance of the battery). What is the percent error caused by
the nonzero resistance of the galvanometer?
First calculate the actual voltage drop. a
R1=10 k
R eq  R1 +R2 =15 103 
V
6V
-3
I


0.4

10
A
3
R eq 15 10 
Vab = IR   0.4 10-3 10 103    4 V
R2=5 k
V=6 V
b
The measurement is made with the galvanometer.
60  and 10 k resistors in parallel
are equivalent to an 59.6  resistor.
The total equivalent resistance is
5059.6 , so 1.19x10-3 A of current
flows from the battery.
The voltage drop from a to b is then
measured to be
6-(1.19x10-3)(5000)=0.07 V.
The percent error is.
4 -.07
% Error =
100 = 98%
4
RG=60 
a
G
R1=10 k
R2=5 k
I=1.19 mA
V=6 V
Your opinions? Would you pay for this voltmeter?
b
To reduce the percent error, the device being used as a
voltmeter must have a very large resistance, so a voltmeter
can be made from galvanometer in series with a large
resistance.
a
V
Vab
b

a
RSer
RG
G
b
Vab
Everything inside the blue box is the voltmeter.
Homework hints: “the galvanometer reads 1A full scale” would mean a current of IG=1A would produce
a full-scale deflection of the galvanometer needle.
If you want the voltmeter shown to read 10V full scale, then the selected RSer must result in IG=1A
when Vab=10V.
Example: a voltmeter of resistance 100 k is used to measure
the voltage drop across a 10 k resistor in series with a 6 V
battery and a 5 k resistor (neglect the internal resistance of
the battery). What is the percent error caused by the nonzero
resistance of the voltmeter?
We already calculated the actual
voltage drop (3 slides back).
Vab = IR   0.4 10-3 10 103    4 V
a
R1=10 k
R2=5 k
V=6 V
b
The measurement is now made with the voltmeter.
100 k and 10 k resistors in
parallel are equivalent to an 9090 
resistor. The total equivalent
resistance is 14090 , so 4.26x10-4
A of current flows from the battery.
The voltage drop from a to b is then
measured to be
6-(4.26x10-4)(5000)=3.9 V.
The percent error is.
4 - 3.9
% Error =
100 = 2.5%
4
RV=100 k
a
V
R1=10 k
R2=5 k
I=.426 mA
V=6 V
Not great, but much better. Larger Rser is needed for high accuracy.
b
Measuring Instruments: Ohmmeter
An ohmmeter measures resistance. An ohmmeter is made
from a galvanometer, a series resistance, and a battery.
RG
RSer
G
Everything inside the blue
box is the ohmmeter.
The ohmmeter is connected in parallel with the unknown
resistance with external power off. The ohmmeter battery
causes current to flow, and Ohm’s law is used to determine
the unknown resistance.

R=?
To measure a really small resistance, an ohmmeter won’t
work.
Solution: four-point probe.
A
V
Measure current and voltage separately, apply Ohm’s law.
reference: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/movcoil.html#c4
Today’s agenda:
Measuring Instruments: ammeter, voltmeter, ohmmeter.
You must be able to calculate currents and voltages in circuits that contain “real”
measuring instruments.
RC Circuits.
You must be able to calculate currents and voltages in circuits containing both a resistor
and a capacitor. You must be able to calculate the time constant of an RC circuit, or use
the time constant in other calculations.
RC Circuits
RC circuits contain both a resistor R and a capacitor C (duh).
Until now we have assumed that charge is
instantly placed on a capacitor by an emf.
Q
t
The approximation resulting from this
assumption is reasonable, provided the
resistance between the emf and the capacitor
being charged/discharged is small.
If the resistance between the emf and the
capacitor is finite, then the capacitor does not
change instantaneously.
Q
t
Charging a Capacitor
Switch open, no current flows.
I
+q +
-q
Close switch, current flows.
q
- IR = 0
C
C
 +-
Apply Kirchoff’s loop rule*
(green loop) at the instant
charge on C is q.
ε-
-
This equation is
deceptively
complex because
I depends on q
and both depend
on time.
R
switch
t<0
t>0
*Convention for capacitors is “like” batteries: negative if going across from + to -.
Limiting Cases
q
ε - - IR = 0
C
When t=0, q=0 and I0=/R.
When t is “large,” the capacitor
is fully charged, the current
“shuts off,” and Q=C.
I
+
-
C
 +R
switch
 = IR is true only at time t=0! VR = IR is always true, but VR is the potential difference across
the resistor, which you may not know. Using V = IR to find the voltage across the capacitor is
likely to lead to mistakes unless you are very careful.
Math:
ε-
q
- IR = 0
C
ε q
I= R RC
dq ε q
Cε q
Cε - q
= =
=
dt R RC RC RC
RC
dq
dt
=
Cε - q RC
dq
dt
=q- Cε
RC
More math:

q
0
t dt
dq
=-
0 RC
q- Cε
1 t
ln  q- Cε  0 = dt

0
RC
q
t
 q- Cε 
ln 
=RC
 -Cε 
t
q- Cε
= e RC
-Cε
q- Cε = -Cε e
-
t
RC
Still more math:
q = Cε - Cε e
-
t
RC
t


RC
q = Cε 1- e 


t


RC
q  t  = Q 1 - e 


Why not just
solve this for
q and I?
ε-
q
- IR = 0
C
dq Cε - RCt
Cε - RCt
ε - RCt
ε - t
It =
=
e =
e = e = e
dt RC
RC
R
R
 = RC is the “time constant” of the circuit; it tells us “how
fast” the capacitor charges and discharges.
recall that this is I0,
also called Imax
Charging a capacitor; summary:
t


RC
q  t  = Qfinal 1- e 


ε - RCt
It = e
R
Charging Capacitor
0.01
0.05
0.008
0.04
0.006
0.03
I (A)
q (C)
Charging Capacitor
0.004
0.002
0.02
0.01
0
0
0
0.2
0.4
0.6
t (s)
0.8
1
0
0.2
0.4
0.6
0.8
t (s)
Sample plots with =10 V, R=200 , and C=1000 F.
RC=0.2 s
1
In a time t=RC, the capacitor charges to Q(1-e-1) or 63% of its
capacity…
…and the current drops to Imax(e-1) or 37% of its maximum.
Charging Capacitor
0.01
0.05
0.008
0.04
0.006
0.03
I (A)
q (C)
Charging Capacitor
0.004
0.002
0.02
0.01
0
0
0
0.2
0.4
0.6
0.8
1
t (s)
0
0.2
0.4
0.6
0.8
t (s)
RC=0.2 s
=RC is called the time constant of the RC circuit
1
Discharging a Capacitor
Capacitor charged, switch
open, no current flows.
Close switch, current flows.
Apply Kirchoff’s loop rule*
(green loop) at the instant
charge on C is q.
q
- IR = 0
C
I
C
+Q
+q
-q
-Q
R
switch
t<0
t>0
*Convention for capacitors is “like” batteries: positive if going across from - to +.
Math:
q
- IR = 0
C
IR =
q
C
I=
dq
dt
-R
dq q
=
dt C
dq
dt
=q
RC
negative because
charge decreases
More math:
t dt
dq
1 t
Q q = - 0 RC = - RC 0 dt
q
1 t
ln  q  Q = dt

0
RC
q
t
q
ln   = RC
Q
q(t) = Q e
-
t
RC
t
dq Q - RCt
I(t) = - =
e = I0 e RC
dt RC
same equation
as for charging
Disharging a capacitor; summary:
q(t) = Q0 e
-
t
RC
I  t  = I0 e
0.01
0.05
0.008
0.04
0.006
0.03
0.004
0.02
0.002
0.01
0
0
0
0.2
0.4
0.6
t (s)
t
RC
Discharging Capacitor
I (A)
q (C)
Discharging Capacitor
-
0.8
1
0
0.2
0.4
0.6
0.8
t (s)
Sample plots with =10 V, R=200 , and C=1000 F.
RC=0.2 s
1
In a time t=RC, the capacitor discharges to Qe-1 or 37% of its
capacity…
…and the current drops to Imax(e-1) or 37% of its maximum.
Discharging Capacitor
0.01
0.05
0.008
0.04
0.006
0.03
I (A)
q (C)
Discharging Capacitor
0.004
0.02
0.002
0.01
0
0
0
0.2
0.4
0.6
0.8
1
t (s)
0
0.2
0.4
0.6
t (s)
RC=0.2 s
0.8
1
Notes
ε - RCt
It = e
R
I  t  = I0 e
-
t
RC
This is for charging a capacitor.
/R = I0 = Imax is the initial current,
and depends on the charging emf
and the resistor.
This is for discharging a capacitor.
I0 = Q/RC, and depends on how
much charge Q the capacitor
started with.
I0 for charging is equal to I0 for discharging only if the
discharging capacitor was fully charged.
Notes
In a series RC circuit, the same current I flows through both
the capacitor and the resistor. Sometimes this fact comes in
handy.
In a series RC circuit, where a source of emf is present (so
this is for capacitor charging problems)...
VR + VC = V
VR = V - VC = IR
V - VC
I =
R
Vc and I must be at the same instant in time for this to work.
Any technique that begins with a starting equation and is worked
correctly is acceptable, but I don’t recommend trying to memorize a
bunch of special cases. Starting with I(t) = dq(t)/dt always works.
Notes
In a discharging capacitor problem...
VR = VC = IR
VC
I=
R
So sometimes you can “get away” with using V = IR, where V
is the potential difference across the capacitor (if the circuit
has only a resistor and a capacitor).
Rather than hoping you get lucky and “get away” with using
V = IR, I recommend you understand the physics of the
circuit!
Homework Hints
Q(t) = CV(t)
This is always true for a capacitor.
t


RC
q  t  = Qfinal 1- e 


Qfinal = C, where  is the potential
difference of the charging emf.
q(t) = Q0 e
V = IR
-
t
RC
Q0 is the charge on the capacitor
at the start of discharge. Q0 = C
only if you let the capacitor charge
for a “long time.”
Ohm’s law applies to resistors, not
capacitors. Can give you the
current only if you know V across
the resistor. Safer to take dq/dt.
Example: For the circuit shown C = 8 μF and ΔV = 30 V.
Initially the capacitor is uncharged. The switch S is then closed
and the capacitor begins to charge. Determine the charge on
the capacitor at time t = 0.693RC, after the switch is closed.
(From a prior test.) Also determine the current through the
capacitor and voltage across the capacitor terminals at that
time.
C
To be worked at the
blackboard in lecture.
R
S
ΔV
Demo
Charging and discharging
a capacitor.
Instead of doing a physical demo, if I have time I will do a virtual demo using the
applet linked on the next slide. The applet illustrates the same principles as the
physical demo.
make your own capacitor circuits
http://phet.colorado.edu/en/simulation/circuit-construction-kit-ac
For a “pre-built” RC circuit that lets you both charge and discharge (through separate switches), download
this file, put it in your “my documents” folder, run the circuit construction applet (link above), maximize it,
then select “load” in the upper right. Click on the “capacitor_circuit” file and give the program permission to
run it. You can put voltmeters and ammeters in your circuit. You can change values or R, C, and V. Also,
click on the “current chart” button for a plot of current (you can have more than one in your applet) or the
“voltage chart” button for a plot of voltage.
more applets
http://webphysics.davidson.edu/physlet_resources/bu_semester2/c11_RC.html
http://subaru.univ-lemans.fr/AccesLibre/UM/Pedago/physique/02/electri/condo2.html
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=31.0