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Oct. 8
19.2-19.4
Oct. 10
boardwork
Quiz 4
Oct. 13
19.5-19.9
Oct. 15
boardwork
Oct. 17
boardwork
Quiz 5
Oct. 20
review
Oct. 22
Exam 2
Ch. 18, 19
Oct. 24
20.1-20.2
Oct. 6
19.1
“Physics is becoming too difficult for the physicists.”—mathematician David
Hilbert, referring to quantum mechanics
19.5 EMF’s in Series and in Parallel: Charging a Battery
If you put batteries in series the “right way,” their voltages add:
+
6V
=
3V
9V
If you put batteries in series the “wrong way,” their voltages
add algebraically:
+
=
magnitudes only 
6V
3V
chosen loop direction
-6 V
+3 V
3V
-3 V  algebraically,
using chosen loop
direction
Algebraic addition of voltages for batteries in series comes
directly from Kirchoff’s loop rule.
This applies to any source of emf, not just batteries!
Why would you want to connect batteries in series?
More voltage! Brighter flashlights, etc. Chemical reactions in
batteries yield a fixed voltage. Without changing the chemical
reaction (i.e., inventing a new battery type), the only way to
change voltage is to connect batteries in series.
Go to www.howstuffworks.com to see how batteries work.
They even expose the secret of the 9 volt battery!
Click on the picture above only if you are mature enough to
handle this graphic exposé.
Go to www.howstuffworks.com to see how batteries work.
They even expose the secret of the 9 volt battery!
Shocking!
Six 1.5 V batteries in series!
Why would you want to connect batteries in series the
“wrong” way?
You probably don’t want to.
Use could use one battery to charge another—doesn’t seem too
useful to me, although might be in special cases.
But remember, Kirchoff’s loop rule applies to all emf’s.
You could connect a source of emf – like the alternator in your
car – so that it charges a battery.
Rechargeable batteries use an ac to dc converter as a source of
emf for recharging.
Could you connect batteries (or sources of emf) in parallel?
Sure!
a
b
3V
3V
You would still have a 3 V voltage drop across your resistor,
but the two batteries in parallel would “last” longer than a
single battery.
You could use Kirchoff’s rules to analyze this circuit and show
that Vab = 3 V.
19.6 Circuits Containing Capacitors in Series and in Parallel
Vab
Capacitor:
C
Capacitors connected in parallel:
C1
a
C2
b
C2
+ V
The voltage drop from a to b must equal V.
Vab = V = voltage drop across each individual capacitor.
C1
OSE: Q = C V
 Q1 = C1 V
& Q2 = C2 V
Q1
a
+
Q2
& Q3 = C3 V
C2 -
C3
Q3
+ V
Now imagine replacing the parallel
combination of capacitors by a single
equivalent capacitor.
By “equivalent,” we mean “stores the same
total charge if the voltage is the same.”
Q1 + Q2 + Q3 = Ceq V = Q
a
Ceq
Q
+ V
Summarizing the equations on the last slide:
Q1 = C1 V
Q2 = C2 V
Q3 = C3 V
C1
C2
a
Q1 + Q2 + Q3 = Ceq V
C2
Using Q1 = C1V, etc., in the second line gives
+ -
C1V + C2V + C3V = Ceq V
C1 + C2 + C3 = Ceq
V
(after dividing both sides by V)
Generalizing:
OSE:
Ceq = Ci
(capacitors in parallel)
Does this remind you of any of our resistor equations?
See Giancoli’s comment on why this makes sense, p. 569.
b
Capacitors connected in series:
C1
C2
C3
+ +Q V -Q
An amount of charge +Q flows from the battery to the left plate
of C1. (Of course, the charge doesn’t all flow at once).
An amount of charge -Q flows from the battery to the right
plate of C3. Note that +Q and –Q must be the same in
magnitude but of opposite sign.
The charges +Q and –Q attract equal and opposite charges to
the other plates of their respective capacitors:
C1
+Q -Q
A
C2
+Q -Q
B
C3
+Q -Q
+ V
These equal and opposite charges came from the originally
neutral circuit regions A and B.
Because region A must be neutral, there must be a charge +Q
on the left plate of C2.
Because region B must be neutral, there must be a charge --Q
on the right plate of C2.
Here’s the circuit after the charges have moved and a steady
state condition has been reached:
a
C1
+Q -Q
V1
C2
A
+Q -Q
V2
C3
B
b
+Q -Q
V3
+ V
The charges on C1, C2, and C3 are the same, and are
Q = C1 V1
Q = C2 V2
Q = C3 V3
But we don’t know V1, V2, and V3 yet.
We do know that Vab = V and also Vab = V1 + V2 + V3.
Let’s replace the three capacitors by a single equivalent
capacitor.
Ceq
+Q -Q
V
+ V
By “equivalent” we mean V is the same as the total voltage
drop across the three capacitors, and the amount of charge Q
that flowed out of the battery is the same as when there were
three capacitors.
Q = Ceq V
Collecting equations:
Q = C1 V1
Q = C2 V2
Q = C3 V3
Vab = V = V1 + V2 + V3.
Q = Ceq V
Substituting for V1, V2, and V3:
Q
Q
Q
V=
+
+
C1 C 2 C 3
Substituting for V:
Q
Q
Q
Q
=
+
+
Ceq C1 C2 C3
Dividing both sides by Q:
1
1
1
1
=
+
+
Ceq C1 C2 C3
Generalizing:
OSE:
1
1
=
Ceq
Ci
i
(capacitors in series)
Does this remind you of any of our resistor equations?
Example 19-9
Determine the capacitance of a
single capacitor that will have the
same effect as the combination
shown. Use C1 = C2 = C3 = C.
C2
C1
C3
I don’t see a series combination of capacitors, but I do see a
parallel combination.
C23 = C2 + C3 = C + C = 2C
Now I see a series combination.
1
1
1
=
+
Ceq C1 C23
1
1
1
2
1
3
= +
=
+
=
Ceq C 2C 2C 2C 2C
C eq
2
= C
3
C23 = 2C
C 1= C
19.7 Circuits Containing A Resistor and a Capacitor
We won’t analyze complex circuits involving many emf’s,
resistors, and capacitors in series and parallel.
We will look at simple circuits involving
a resistor and a capacitor together.
Such a circuit is called an RC circuit.
(The circuit to the right is the same as
the one in Fig. 19-17.)
R
C
- +

S
When switch S is closed, current
flows.
The voltage across the capacitor will
eventually equal the battery
voltage.
However, the flow of current is not
instantaneous, but takes time.
R
- +
-Q +Q

C
S
As a result, the voltage across the capacitor increases with
time.
OSE :
V =ε 1- e

- t / RC

This equation can easily be derived using
a bit of calculus. Ask if you want to see
the derivation.
V = ε 1 - e- t / RC


The plot to the right shows
the voltage vs. time for a 1
F capacitor in series with a
10  resistor and a 10 V
battery.
t = RC
Will the voltage across the
capacitor “ever” reach 10 V?
Note the RC in the exponent in the equation. RC has units of
time and is called the “time constant” of the circuit. When t =
RC, V =  (1 – 1/e) = 0.63 . The voltage across the
capacitor reaches 63% of the battery voltage within a time RC.
In the circuit for the graph, RC = (10)(10-6) = 10-5 = 10 s.
The plot to the right
shows the voltage vs.
time for C = 1 F and
 = 10V. For the blue
curve, R = 10 and
for the “violet” curve,
R = 1.
Notice how much
more rapidly the
capacitor charges in
the circuit with smaller
RC.
Why did we jump to an RC circuit and not study the time
behavior of a circuit with only a capacitor?
All circuits have some resistance, if only in the wires and the
emf source. Therefore “all” circuits with capacitors are RC
circuits.
R
If the battery is removed from our
circuit, the capacitor discharges through
the resistor.
You can show that the voltage across
the capacitor decays exponentially:
OSE :
V = V0 e- t / RC
C
-Q +Q
S
The plot to the right
shows the voltage vs.
time for our circuit.
R = 1 ; RC = 1 s
R = 10 ; RC = 10 s
Example 19-10
If a charged capacitor, C = 35 F, is connected to a resistance,
R = 120 , how much time will elapse until the voltage falls to
10 percent of its original (maximum) value?
OSE :
V = V0 e- t / RC
After a time t1 the voltage drops to a value V1:
t1 is what we want
to solve for
V1 = V0 e- t1 / RC
V1
= e- t1 / RC
V0
 V1 
ln   = ln e- t1 / RC
 V0 


 V1 
ln   = ln e- t1 / RC
 V0 


copied from
previous page
 V1 
ln   = - t1 / RC
 V0 
 V1 
t1 = - RC ln  
 V0 

t1 = - 120  35×10
-6

 0.1 V0 
ln 

 V0 
t1 = 9.7 ×10-3 s
19.8 Heart Pacemakers
Please read this section!
19.9 Electric Hazards: Leakage Current
If you value your life, you will read this section!
Handy stuff to cut and paste:
C
R
+ V