3/17/2009
PHYS202
– SPRING
2009
Batteries
• A   battery   is   a   device   that   provides   a   potential   difference   to   two   terminals.
• Different   metals   in   an   electrolyte   will   create   a   potential   difference,   and   maintain   the   difference   so   long   as   the   load on   the   battery   is   not   to   large.
• Chemical   potential   energy   is   converted   to   electric   potential   energy.
• The   battery   is   said   to   provide   Electromotive   Force ( E )
1

Ohm’s   Law
• Let   us   consider   a   circuit .
• Connections   are   shown   as   solid   lines.
• Batteries   are   shown   as   two   lines   perpendicular   to   connections   with   one   shorter   than   the   other.
• The   conventional   current   flows   from   the   positive   to   negative   terminals.
Positive
I  ‐ current
Negative
V R  ‐ resistance
Ohm’s   Law
• Normally   current   is   not   labeled .
• The   current   flow   is   proportional   to   the   voltage:
V ∝ I
•
•
Resistance   is   the   constant   of   proportionality:   Ohm’s   Law
V = IR
• An   element   within   an   electric   circuit   with   resistance   is   called   a   resistor.
• Ohm’s   Law   is   not   a   fundamental   rule   of   nature.
• It   isn’t   true   in   every   case,   everywhere   – but   it   is   a   very   useful   rule   in   practical   situations.
I  ‐ current
V
R  ‐ resistance
3/17/2009
2

Example
20.4
– Suppose   that   the   resistance   between   the   walls   of   a   biological   cell   is   5.0
x   10 9 Ω .
(a)   What   is   the   current   when   the   potential   difference   between   the   walls   is   75   mV?
Use   Ohm’s   law:
I =
V = IR ⇒ I =
V
R
0.075V
× 9 Ω
= 1.5 10 − 11 A
(b)   If   the   current   is   composed   of   Na + ions   (q   =   +e),   how   many   such   ions   flow   in
0.50
s?
I =
Δ Q
Δ t
⇒ Q I t
N =
+
N Q
1Na
1 e
(
1.5 10 − 11
)
⇒ N =
Na
+ = × e
1.602 10 − 19 C
Na +
+ ions
Resistance
• Resistance   is   given   in   terms   of   an   objects   tendency   to   resist   the   flow   of   electrons   ρ .
• Resistance   increases   with   the   length of   the   material   and   decreases   with   its   cross   sectional   area .
R =
A
L
• All   materials   have   some   resistivity,   ρ , to   differing   degrees.
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3

Resistance
• As   expected,   insulators   have   very   high   resistivity   and   conductors   very   low   resistivity.
• Semi ‐ conductors   fall   in   between   and   their   resistivity   depends   heavily   on   the   amount   of   impurities.
Resistance   and   Temperature
• As   temperature   changes,   so   does   resistivity.
• In   small   temperature   ranges,   this   is   given   by
ρ ρ
0
⎡ 1 + α ( −
0
)
• Where   the   ρ
0 is   the   resistivity   at   temperature   T
0
, and α is   the   temperature   coefficient   of   resistivity .
3/17/2009
4

Example
• Problem   20.14
– A   copper   wire   has   a   resistance   of   38.0
Ω at
25   ˚ C   and   43.7
Ω at   55   ˚ C.
What   is   the   temperature   coefficient   of   resistivity?
• Solving
ρ ρ 1 + α (
T T
0
) for   α
α =
R
–1
R
0
–
0
=
43.7
Ω
38.0
Ω
–1
=
–1
P =
Δ
Electric   Power
Energy
⇒ P =
Δ qV
=
Δ t Δ t
Δ q
Δ t
V
The   amount   of   power   used   in   a   portion   of   a   circuit,   is   the   product   of   the   voltage   drop   and   the   current   in   that   section   of   the   circuit.
Using   Ohm’s   Law:                ,
=
P
R
V
P =
V 2
R
3/17/2009
5

Example
20.26
– A   piece   of   Nichrome wire   has   a   radius   of   6.5
x   10 ‐ 4 m.
It   is   used   in   a   laboratory   to   make   a   heater   that   uses    4.00
x   10 2 W   of   power   when   connected   to   a   voltage   source   of   120   V.
Ignoring   the   effect   of   temperature   on   resistance,   estimate   the   necessary   length   of   wire.
Looking   up   the   resistance   of   NiChrome wire,   we   find   ρ =   100   x   10 ‐ 8 Ω∙ m.
We   need   to   find   the   resistance   of   a   wire   with   a   particular   radius:
R =
ρ
A
L
=
π
ρ r
L
2
We   need   the   resistance   needed   to   dissipate   400W   by   a   voltage   of   120V:
P =
V
R
2
⇒ R =
V
P
2
Thus,   the   length   of   the   wire   must   be
ρ
π r
L
2
=
V
P
2
⇒
L =
(
120 V
) 2 π
(
(
100 10 − 8 Ω ⋅
L =
V 2 π r 2
ρ P
) (
− 4
) 2
)
= 48 m
Example
20.27
– Tungsten   has   a   temperature   coefficient   of   resistivity   of   0.0045
(C°) − 1 .
A   tungsten   wire   is   connected   to   a   source   of   constant   voltage   via   a   switch.
At   the   instant   the   switch   is   closed,   the   temperature   of   the   wire   is   28°C,   and   the   initial   power   delivered   to   the   wire   is   P
0
.
At   what   wire   temperature   will   the   power   that   is   delivered   to   the   wire   be   decreased   to    ½   of   P
0
?
If   the   source   of   the   voltage   is   constant   and   temperature   affects   resistance,   can compe the   power   we   find P
0
=
V
R
2 and
1
2
P
0
=
V
2
2
R
,   thus   what   we   are   looking   for   is   the   point   where   the   resistance   is   twice   the   value   at   28 ˚ C.
Since   the   diameter   and   length   of   the   wire   are   unchanged,   we   are   simply   looking   for   where   the   resistivity   is   doubled:
2 ρ
0
= ρ
0
⎣
1 + α ( −
0
) ⇒ α ( −
0
) = 1
T
1
α
T
0
=
1
( ) − 1
+
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6

Alternating   Current
• Let   us   treat   the   voltage   source   as   oscillating as   a   sine   wave   (see   image   to   right)
• Normal   AC   voltage   in   the   US   has   a frequency   of   60   Hz   (e.g.
most   of   Europe   is
50   Hz).
• So   what   is   the   current?
‐ Ohm’s   Law   gives
I =
V
=
V
R R
0
I = I
0
( π ft
(
)
π ft
)
• The   current   oscillates   in   the   same   way   as   the   voltage.
=
0
( π ft
) f = 60 Hz
• Power   is
( π ft
)
•
Average   power   is   half   of   the peak   power.
AC   Power
• Average   Power   is
P =
2
= ⎜
⎛
⎝
V
0
2
⎞⎛
⎠⎝
I
0
2
⎞
⎠
Average
Power
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7

AC   Power
• We   define   this   level   of   voltage   and   current   as   the   “root   mean   square”   (rms)   voltage   and   current.
V rms
=
V
0
2
I rms
=
I
0
2
• Since   this   is   the   average   power   delivered   over   one   period,   it   equates   to   the   power   delivered   in   an   AC   circuit.
• We   apply   Ohm’s   law    V   =   IR
=
P =
V 2 rms
R rms
• In   the   US,   we   have   V rms
=   120V    (240V maximum   voltage   of   170V   (340V)
for   heavy   duty),   which   equates   to   a
Example
20.36
– A   light   bulb   is   connected   to   a   120.0V
wall   socket.
The   current   in   the   bulb   depends   on   the   time   t according   to   the   relation   I =   (0.707
A)   sin   [(314   Hz)t].
(a)   What   is   the   frequency   of   the   alternating   current?
We   know   that   I =   I
0 sin(2 π ft ),   so,   f =
314 Hz
2 π
= 50 Hz
(b)   Determine   the   resistance   of   the   bulb’s   filament.
Using   Ohm’s   law, V rms
=
R =
⇒ R =
V rm s
I rms
( )
= 240 Ω
0.707 A
=
V
I
0
/ 2
(c)   What   is   the   average   power   delivered   to   the   light   bulb?
rms rm s
=
V r
2 m s
R
=
(
120 V
240 Ω
) 2
= 60 W
3/17/2009
8

AC   Power   Transmission
• Note   that   power   transmitted   through   a   power   line   is
=
P =
V 2 rms
R rms
•
• Let   us   say   that   10   km   of   wire   has   a   resistance   of   1.7
Ω .
If   a   power   line   is   to   transmit   10kW,   is   it   better   to   have   high   voltage   or   high   current?
Voltage Current Power loss
When power
dealing loss
with through wire,   use   P =   I 2 R.
a
1   kV
10kV
100V
10A
1A
100A
170W     (1.7%)
1.7W
(0.017%)
1.7kW
(17%)
Power   loss   in   a   wire
• Current   must   stay   constant   through   a   wire,   or   charge   would   start   to   build   up.
• The   Voltage   drop   is   continuous,   so   we   would   have   to   calculate   how   much   power   is   lost   at   each   point   in   the   wire   due   to   the   constant   drop   in   voltage.
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9

Wiring   Resistors   in   Series
• Take   two   resistors   ( R
1 and   R
2
)   and   place   them   in   series.
V
1
V
2
R
1
R
2
V
• Since   charge   does   not   build   up,   the   current   through   both   must   be   the   same,   so   each   must   experience   a   voltage   drop:
= +
2
= IR
1
+ IR
2
= (
1
+ R
2
)
IR
Series
= (
1
+ R
2
) ⇒ R
Series
= (
R
1
+ R
2
)
• Resistors   in   Series   add   their   resistances   together   to   get   the   equivalent   resistance.
R
Series
= R
1
+ R
2
+
Wiring   Resistors   in   Series
• Power   for   resistors   in   series
• Find   the   current   in   the   circuit: I =
V
R
1
+ R
2
• The   corresponding   voltage   drops across   each   resistor   are: V
1
= IR
1
• The   power   dissipated   is P
1
= I R
1
P
1
=
V 2
(
R
1
+ R
2
) 2
R
1 and and and
V
V
2
= IR
2
P
2
= I R
2
P
2
=
V 2
(
R
1
+ R
2
) 2
R
2
V
1
R
1
• The   total   power   dissipated   is   as   expected
= +
2
=
V 2
(
R
1
+ R
2
V 2
) 2
R
1
+
(
R
1
+ R
2
) 2
R
2
⇒ P =
V 2
R
1
+ R
2
= VI
V
2
R
2
3/17/2009
10

Examples
20.42
– The   current   in   a    47 Ω resistor   is   0.12
A.
This   resistor   is   in   series   with   a    28 Ω resistor,   and   the   series   combination   is   connected   across   a   battery.
What   is   the   battery   voltage?
For   resistors   in   series,   the   current   must   be   the   same,   so,
V = (
1
+ R
2
) ( )(
28
)
V = 9 V
3/17/2009
Examples
20.47
– Two   resistances,   R
1 and   R
2
,   are   connected   in   series   across   a   12 ‐ V   battery.
The   current   increases   by   0.20
A   when   R
2 is   removed,   leaving   R
1 battery.
However,   the   current   increases   by   just   0.10
A   when   R
1
R
2 connected   across   the   battery.
Find   (a)   R
1 and   (b)   R
2
.
connected is
across removed,
the   leaving
Examining   the   series   situation,   we   find   that   I
V =
(
1
+
1
R
2
=   I
) s
+0.20A
and   I
2
=   I s
+0.10A:
V = I R =
(
I s
+ 0.20 A
)
R
1
V = I R
2
=
(
I s
+ 0.10 A
)
R
2
Thus,
V = I s
⎡
⎢
(
I s
V
+ 0.
20 A
R
1
=
(
I s
+
V
0.20 A
)
R
2
=
(
I s
V
+ 0.10 A
)
V
+ 0.10 A
)
⎤
⎥
I s
I s
2 + (
0.30 A
)
I s
+
⇒
I s
2
(
I s
+
= 0.020
A
0.10 A
)(
I s
I s
2
+ 0.20 A
) = 2 I s
2
+ (
0.30 A
)
I s
+ (
0.30 A
)
I s
11

Examples
20.47
– (cont.)
Thus,                         ,   so   substituting   back,
V = (
I s
+ 0.20 A
)
R
1
R
1
=
I s
+
V
0.20 A
=
12 V
+
R
1
V = (
I s
+ 0.10 A
)
R
2
R
2
=
I s
+
V
0.10 A
=
12 V
+
R
2
Note   that   this   is   not   the   only   method   to   get   to   the   solutions.
Wiring   Resistors   in   Parallel
• Take   two   resistors   ( R
1 and   R
2
)   and   place   them   in   parallel.
V R
1
R
2
• The   battery   provides   the   same   voltage   to   each   resistor:
V R
1
R
2
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12

Wiring   Resistors   in   Parallel
• Take   two   resistors   ( R
1 them   in   parallel.
and   R
2
)   and   place
V I
1
• The   battery   provides   the   same   voltage   to each   resistor.
• The   currents   must   add   up   to   the   total   current   through   the   battery,
R
1
R
V parallel
I I
1
I
2
=
V
R
1
+
V
R
2
= V ⎜
⎛
⎝
1
+
1
R
1
R
2
⎟
⎞
⎠
= V ⎜
⎛
⎝
1
+
1
R
1
R
2
⎟
⎞
⎠
⇒
R
1 parallel
=
1
+
1
R
1
R
2
I
2
R
2
• For   resistors   in   parallel,   the   net   resistance   is   the   inverse   of   the   sum   of   the   inverse   of   the   resistances.
R
1 parallel
=
1
+
1
R
1
R
2
+ "
Wiring   Resistors   in   Parallel
• Power   for   resistors   in   parallel:
• Find   the   current   through   the resistors:
I
1
=
V
R
1
I
2
=
V
R
2
V I
1
• Power   is ⇒ P
1
= I V
• The   total   power   given   from   the   battery   is
P
1
=
V 2
R
1
P
2
P
2
= I V
=
V 2
R
2
= +
2
=
V
R
2
1
+
V
R
2
2
= V 2
⎛
⎝
1
+
1
R
1
R
2
⎞
⎠
=
V 2
R parallel
R
1
I
2
R
2
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13

Examples
20.52
– The   drawing   shows   a   circuit   that   contains a   battery,   two   resistors,   and   a   switch.
What   is the   equivalent   resistance   of   the   circuit   when the   switch   is   (a)   open   and   (b)   closed?
When   the   switch   is   open,   no   current   flows   through   R2,   thus   the   equivalent   resistance   is   65.0
Ω .
When   the   switch   is   closed   the   resistors   are   in   parallel,   so   the   equivalent   resistance   is, 1
R parallel
=
1
+
1
R
1
R
2
R paralle l
=
(
65.0
Ω
⇒
)(
96.0
R parallel
Ω
65.0
Ω + 96.0
Ω
)
=
=
R
1
38.8
Ω
+ R
2
What   is   the   total   power   delivered   to   the   resistors   when   the   switch   is   (c)   open   and
(d)   closed?
P op
V 2
= =
R
1
(
9.00 V
65.0
Ω
) 2
= 1.25 W
P cl
V 2 V 2
= + =
R
1
R
2
(
9.00 V
65.0
Ω
+
9.00 V
96.0
Ω
) 2
= 2.09 W
Examples
20.56
– A   cylindrical   aluminum   pipe   of   length   1.50
m   has   an   inner   radius   of    2.00
x   10 ‐ 3 m   and   an   outer   radius   of   3.00
x   10 ‐ 3 m.
The   interior   of   the   pipe   is   completely   filled   with   copper.
What   is   the   resistance   of   this   unit?
As   the   hint   in   the   book   suggests,   we   can   treat   this   as   two   parallel   resistors.
We   need   to   find   the   resistance   of   each.
The   cross   sectional   areas   are
A
Cu
= π
Thus   the   resistances,
R in
2
R
Cu
=
ρ
Cu
L
A
Cu
=
ρ
Cu
π R 2
L in
A
Al
R
Al
=
= π
(
ρ
Al
L
A
Al
R 2 out
− R in
2
=
π
ρ
Al
)
L
(
R 2 out
− R 2 in
)
Placing   the   resistances   in   parallel ,
1
R net
1
R net
=
=
1
R
Cu
1
R
Cu
+
+
1
R
Al
1
R
Al
=
=
π
ρ
R 2 in
Cu
L
πρ R 2
Al in
+
+
π
(
πρ
Cu
R 2 out
− R 2 in
ρ
Al
L
(
)
R 2 out
− R 2 in
ρ ρ
Al
L
)
R net
=
R 2
Al in
+ ρ
Cu
Al
(
L
R 2 out
− R 2 in
) ) = − 3 Ω
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14

V
Resistors   in   Series   and   Parallel
• Circuits   can   be   wired   with   resistors   in   series   and   parallel.
• To   find   the   current   and   power   delivered   by   a   battery,   create   equivalent   circuits   by   combining   resistors:
• Each   circuit   is   equivalent   from   the   point   of   view   of   the   battery.
• Deal   with   each   part   of   the   circuit   where   you   are   combining   two   resistors,   then   move   to   the   next.
R
2
R
1
R
3
V R
1
R
3
V
R p
R
S
= R
2
+ R
3
1
=
1
+
1
R p
R
1
R
S
Examples
20.58
– Find   the   equivalent   resistance   between   points
A   and   B   in   the   drawing.
We   must   combine   the   resistors   until   only   one   remains.
Starting   with   the   three   resistors   in   series   on   the   right, we   find   three   resistors   in   series   that   can   be   combined:
2.00 Ω 6.00 Ω
4.00 Ω 3.00 6.00 Ω
R
S
= 1.00
Ω + 2.00
Ω + 3.00
Ω = 6.00
Ω
Now,   there   are   two   resistors   in   parallel   on   the   right,   combining   those   we   get,
2.00 Ω 6.00 Ω
4.00 Ω 2.00 Ω
1
R p
=
1
3.00
Ω
+
1
6.00
Ω
⇒ R p
= 2.00
Ω
3/17/2009
15

Examples
20.58
– (cont.)
Now,   we   have   a   pair   of   resistors   in   series,   combining,
2.00 Ω
4.00 Ω
8.00 Ω
R
S
= 6.00
Ω + 2.00
Ω = 8.00
Ω
We   now   combine   the   two   resistors   in   parallel,
2.00 Ω
2.67 Ω
1
R p
=
1
4.00
Ω
+
1
8.
00 Ω
We   can   now   combine   the   final   resistors   which   are   in   series,
⇒ R p
= 2.667
Ω
R
S
= 2.00
Ω + 2.667
Ω = 4.67
Ω
Examples
20.66
– Three   identical   resistors   are   connected   in   parallel.
The   equivalent   resistance   increases   by    700 Ω when   one   resistor   is   removed   and   connected   in   series   with   the   remaining   two,   which   are   still   in   parallel.
Find   the   resistance   of   each   resistor.
Let   us   define   the   resistance   in   parallel
1 1 1 1
R
1
R R R
=
3
R
⇒ R
1
=
1
3
R
Removing   one   resistor   and   placing   it   in   series   with   the   other   two:
R
1 p
=
1
+
1
=
R
2
R R
⇒ R
2
=
1
+ =
2
R R
3
2
R
R
2
= R
1
+ 7 0 0 Ω
Substituting   our   equations   for   R
1
3
2
R =
1
3
R + 700 Ω and   R
2
⇒ into   the   above,
7
6
R = 700 Ω
R = 600 Ω
3/17/2009
16

Internal   Resistance   of   a   Battery
• One   way   a   battery   is   rated   is   by   its   voltage   between   its   terminals.
• However,   when   a   battery   is   connected   to   a   load   and   there   is   current   flowing,   there   is   some   internal   resistance   to   a   battery.
• The   net   result   is   to   reduce   the   voltage   between   the   terminals.
•
• The   resistance   inside   the   battery   is   constant,   but   the   voltage   drop   of   the   terminals   depends   on   the   load:
V = ( + int
) V term
= V
(
R
+ int
)
The   voltage   across   the   terminals   is   equivalent   to   the   voltage   over   resistor   R.
+
R
V
‐
Terminal   Voltage   and
Internal   resistance
For   internal   resistance   of    0.01
Ω
3/17/2009
17

Terminal   Voltage   and
Current   Through   the   Battery
• The   more   current   through   the   battery,   the   lower   the   terminal   voltage.
• If   a   12V   battery   has   an   internal   resistance   of   0.010
Ω and   a   current   of   10A   flowing   through   the   battery.
The   terminal   drop   is
V term
= − int
V term
=
= 11.9 V
( Ω )
• If   the   current   is   100A,   then
V term
= − int
V term
=
= 11.0 V
( Ω )
3/17/2009
Kirchhoff’s   Rules
• Sometimes   circuits   are   too   complex   to   analyze   by   reducing   parallel   and   series   resistors.
• For   this   we   use   two   concepts:
• 1.
Conservation   of   charge   implies   conservation   of   current.
€ For   conservation   of   current,   the   current   flowing   into   a   junction   must   equal the   current   flowing   out.
• Let   us   assume   that   the   current flows   in   the   circuit   as   drawn.
18

Kirchhoff’s   Rules
• Take   junction   A.
The   current   in   must   equal   the   current   out.
A
I left
I right
•
• Label   each   branch   of   current.
Find
Point   the
Point   A:
B:
currents each   point:
I right
I middle
=
in   an
I middle
+ I left
=
+ out
I left
I right
of
• Note   that   both   equations   in   this   case   are   equivalent.
B
€ For   conservation   of   current,   the   current   flowing   into   a   junction   must   equal the   current   flowing   out.
Kirchhoff’s   Rules
I right
= I middle
+ I left
• 2.
Next   we   consider   loops.
A
I left
I right
• Here   there   are   two   loops.
B
• Just   like   walking   up   and   down   hills,    if   you   go   in   a   loop   the   place   you   start/end   must   be   at   the   same   elevation.
• So,   if   you   make   a   loop   in   a   circuit,   you   come   back   to   the   same   voltage
(potential).
€ This   is   Kirchhoff’s   Loop   rule.
The   sum   of   potential   rises   and   drops   around   any   loop   must   be   zero.
3/17/2009
19

Kirchhoff’s   Rules
I right
= I middle
+ I left
• Go   around   each   loop   noting   the   potential   rises   and   falls.
A
I left
I right
• Left   loop:    Let’s   start   at   point   A   and   go   counter ‐ clockwise:
5V − I left
( )
I left
( )
0
• Right   loop,   again   starting   at   A,   but   let’s   go   clockwise:
(
I right
) ( ) (
12 V
) = 0
• If   we   went   counter ‐ clockwise,   we   would   get   the   same   answer!
12 V − I right
( )
0
B
€ This   is   Kirchhoff’s   Loop   rule.
The   sum   of   potential   rises   and   drops   around   any   loop   must   be   zero.
Kirchhoff’s   Rules
5V −
I
I right left
=
( )
I middle
3 Ω − I left
+
2
I left
( )
0
12 V − I right
0
• We   can   now   find   all   the   currents   as   we   have   three   equations   and   three   unknowns.
I left
€
€
€
I right
I left
I mid
=
=
3.0A
1.0A
=   2.0A
• This   is   an   especially   easy   case   as   there   is   only   one   unknown   in   each   loop   equation.
A
B
I right
€ Because   the   current   is   positive,   it   moves   in   the   direction   assumed.
3/17/2009
20

Examples
20.73
– Two   batteries,   each   with   an   internal   resistance   of
0.015
Ω ,   are   connected   as   in   the   drawing.
In   effect,   the
9.0
‐ V   battery   is   being   used   to   charge   the   8.0
‐ V   battery.
What   is   the   current   in   the   circuit?
For   this   we   only   need   to   apply   Kirchhoff’s   Loop   Rule.
Taking   the   current   as   flowing   in   the   counter ‐ clockwise direction   and   starting   with   the   upper   right   corner:
8.0 V − I
(
0.015
Ω − 0.015
)
− − I = 0
I = − 33A
The   negative   sign   means   that   our   original   assumption   about   the   current   direction   is   wrong.
So   we   have   33A   going   clockwise .
Examples
20.78
– Find   the   current   in   the   4.00
Ω resistor   in   the   drawing.
Specify   the   direction   of   the   current.
We   have   three   loops   (one   of   which   is   the   sum   of   the other   two),   and   three   branches   with   two   junctions   (one of   which   is   redundent).
Let   us   label   the   currents   as   in   the lower   drawing.
Thus   the   junction   rule   at   B   gives,
I
1
I
3
I
2
A
+
The   loop   rules   for   the   left   loop   starting   a   D   and   going clockwise   gives,
3.00
V
3V − I
1
(
2.00
)
6 V − − I 4.00
)
0
The   loop   rules   for   the   left   loop   starting   a   E   and   going clockwise   gives,
− 9 V − I
3
(
4.00
)
6 V − I
2
(
8.00
)
0
D
+
2.00
Ω
I
1
I
3
+
B
C
+
8.00
Ω
6.00
V
I
2
+
4.00
Ω
+
E
F
9.00
V
3/17/2009
21

Examples
2.00
Ω
+
A
20.78
– (cont.)   I
1
I I
2
− I
1
+ + I
3
= 0
(1)
(2) 3.00
V
+
I
1
9 4 I
3
6 8 I
2
= 0 (3)
We
We
have are
three only
equations required   to
and solve
three for   I
3
unknowns.
D
,   so,   substituting   (1)   into   (3),
9 4 I
3
6 8
(
I
1
+ I
3
)
= 0 ⇒ + I
3
+ 8 I
1
= 0
I
3
+
B
C
+
8.00
Ω
6.00
V
+
4.00
Ω
I
2
Multiply   equation   (2)   by   four, 36 8 I + 16 I
3
= 0 ,    then   adding   to   the   above   eqn.,
+ I
3
= 0
I
3
= − 1.82 A
+
E
F
9.00
V
The   negative   sign   indicates   that   the   current   is   in   the   opposite   direction   from   our   initial   assumption.
Examples
20.104
– For   the   circuit   shown   in   the   drawing,   find   the   current   I through   the   2.00
Ω  resistor   and   the   voltage   V of   the   battery   to the   left   of   this   resistor.
Let   us   define   a   current   I
1 flowing shown.
The   junction   rule   gives,
through   the   middle   wire   as
3.00 A
The   top   loop   gives   (going   counter ‐ clockwise),
24 V − I
1
(
6.00
) ( )( ) (
24 6 I
1
− 36 0
I
1
= − 2.00 A
Substituting   into   the   loop   rule   equation   we   get
I = 5.00 A
)( )
0
I
1
3/17/2009
22

Wheatstone   Bridge
• Used   to   find   resistances   precisely.
R
3
• R x is   unknown   but   R
1
,   R
2 accurately   known.
and   R
3 are   A
• R
3 is   a   variable   resistor.
• Method   is   to   vary   R
3 until   the   galvanometer does   not   deflect   (i.e.
points   B and   D are   at
the   same   voltage.
R
1
B
G
D
V
• At   that   time,   the   voltage   drops   across   R
3 and   R
1 are   equal:   I
1
R
1
= I
3
R
3
R x
R
2
C
Wheatstone   Bridge
• I
1
R
1
= I
3
R
3
B
R
3
R x
• So,   it   must   be   true   that   I
2
R
2
= I x
R x
• Since   there   is   no   voltage   between   B   and   D,
I
1
= I
2 and    I
3
= I x
.
A
•
• I
1
R
2
= I
3
R x
• Let’s   divide   the   two   equations.
=
I R I R x
⇒
R
1 =
R
R
2
R
3 x
Or,
R x
=
R
2
R
1
R
3
R
1
G
D
R
2
R
3 is   very
V sensitive, the   galvanometer
so   switch   is   closed   only   briefly   to   check   if   the   current   is   zero.
C
3/17/2009
23

How   does   the
Galvanometer   Work?
• A   DC   Galvanometer   measures   current   by   using   a   coil   spring,   magnets   and   a   loop   of   wire   (more   when   we   reach   chapter   21).
• The   rotation   of   the   loop   of   wire   gives   the   current   passing   through   the   system   on   some   calibrated   scale.
• The   full   scale   current   sensitivity   of   a   galvanometer   is   the   current   required   to   move   the   needle   (i.e.
coil)   deflect   full   scale.
• Depending   on   how   the   galvanometer   is   configured   with   resistors   determines   whether   it   reads   current   or   voltage.
Measuring   Current  ‐ Ammeter
• Galvanometers   can   measure   small   currents.
If   the   sensitivity   is
I m
,   it   can   read   currents   from   about
μ A   then   the    scale   is   1  μ A   to   50  μ A).
0.02
I m up   to   I m
.
(e.g.
if   I m
=   50
• Galvanometers   (like   batteries)   have   small   internal   resistances.
• If   we   want   to   measure   current   on   a   larger   scale,   a    galvanometer   is   placed   is   parallel   with   a   shunt resistor.
• Since   the   voltage   drop   across   the   galvanometer   and   resistor   are   the   same:
G
R
G G
3/17/2009
24

Measuring   Current  ‐ Ammeter
R =
G G
R
G
• So,
I R
G
I
R
• If   the   resistance   of   the   galvanometer   is   known,   then   we   can   set   R to   be   whatever   resistance   needed   to   make   I
R   the   scale   of   the   ammeter.
• For   example,   if   we   want   to   measure   up   to   1   A,   but   the   galvanometer   can   only   take   50  μ A
0.999950
A.
and
R
has
=
(
an   internal   resistance
) (
0.999950 A
Ω of
)
30 Ω ,   then   we   need
= 1.5 10 − 3 Ω
I
R to   be
Measuring   Voltage  ‐ Voltmeter
• A   voltmeter   also   consists   of   a   galvanometer   and   a   resistor   (usually   large),   but   in   series.
G
R
• Imagine   we   want   to   measure   a   full   scale   of   V ,   then   we   would   have
V = I m
( +
G
)
R =
V
−
I m
R
G
• For   example,   if   we   want   the   full   scale   to   be   15V   (again   let’s   take   I m
=   50  μ A   and   R
G
=   30 Ω .
R =
15V
30 300 k Ω
• The   internal   resistance   of   the   galvanometer   is   negligible.
3/17/2009
25

Examples
20.84
– Two   scales   on   a   voltmeter   measure   voltages   up   to   20.0
and   30.0
V,   respectively.
The   resistance   connected   in   series   with   the   galvanometer   is   1680 Ω for   the   20.0
‐ V   scale   and   2930 Ω for   the   30.0
‐ V   scale.
Determine   the   coil   resistance   and   the   full ‐ scale   current   of   the   galvanometer   that   is   used   in   the   voltmeter.
Using   Ohm’s   Law,
V
1
= I m
(
R
1
+ R
G
)
V
2
= I m
(
R
2
+ R
G
)
Subtracting   one   equation   from   the   other,
V
2
V I R
1
− I R
2
Solving   for   I m
,
I m
=
V
2
− V
1
R
1
− R
2
I m
=
2930 Ω −
= 0.008 A
1680 Ω
Substituting   back   into   either   equation   yields   R
G
=   820 Ω
Examples
20.107
– A   galvanometer   with   a   coil   resistance   of   12.0
Ω and   a   full ‐ scale   current   of
0.150
mA is   used   with   a   shunt   resistor   to   make   an   ammeter.
The   ammeter   registers   a   maximum   current   of   4.00
mA.
Find   the   equivalent   resistance   of   the   ammeter.
Remember   that   the   net   current,   is   the   current   through   the   ammeter   and the   galvanometer:
I = I s
+ I
G
⇒ I s
= − I
G
= 3.85mA
The   voltage   across   the   galvanometer   and   shunt   resistor   is   the   same,
I R s
= I R
G
− 3
R s
=
I
G
I s
R
G
=
− 3
(
12 Ω )
R s
= 0.468
Ω
The   equivalent   resistance   is   that   of   the   shunt   resistor   and   galvanometer   coil   in   parallel :
R
1 eq
=
1
R s
+
R
1
G
⇒ R eq
= 0.450
Ω
3/17/2009
26

Capacitors   as elements   in   a   Circuit
• A   Capacitor   can   be   an   element   in   a   circuit,   like   a   battery   or   resistor.
It   is   denoted   as               or
• Since,   q =   CV ,   the   voltage   of   a   capacitor   in   a   circuit   depends   on   the   capacitance   and   the   charge:
V = q
C
• So,   how   does   it   work   in   a   circuit?
Capacitors   in   Parallel and   in   Series
• Just   as   for   resistors,   when   capacitors   are   in   parallel,   they   see   the   same   voltage,   so   the capacitors   will   charge:
V C
1
C
2
C
3 q
1
+ q
2
+ q
3
=
1
+ +
• So,   we   get, q tot
= C V = (
C
1
+ C
2
+ )
V C p
• For   capacitors   in   parallel:
C
P
= (
C
1
+ C
2
+ C
3
)
3/17/2009
27

Capacitors   in   Parallel and   in   Series
• Just   as   for   resistors,   when   capacitors   are   in   series,   they   sum   of   the   voltage   drops   must   equal the   net   voltage   of   the   battery.
So   the capacitors   will   charge:
V C
1
C
2
C
3
= +
2
+ V
3
• Since,   the   wire   in   between   the   capacitors   begins   as   neutral,   the   charge   must   split   evenly,   so   all   capacitors   are   charged   to   the   same   level.
V C
S q
= q
+ q
+ q
C
S
C
1
C
2
C
3
1
=
1
+
1
+
1
C tot
C
1
C
2
C
3
Capacitors   in   Parallel and   in   Series
• The   rules   for   adding   capacitors   and   for   adding   resistors   are   reversed:
Series
Parallel
Resistors
R
S
= R
1
+ R
2
+ R
3
+ "
Capacitors
1
=
1
+
1
+
1
+ "
C
S
C
1
C
2
C
3
1
=
1
+
1
+
1
+ "
R
P
R
1
R
2
R
3
C
S
= C
1
+ C
2
+ C
3
+ "
3/17/2009
28

Charging   and   Discharging
Capacitors   in   a   Circuit
• If   a   Direct   Current   circuit   contains   a   battery,    resistor,    an   initially   uncharged capacitor   and   a   switch,   then   the   capacitor   will   take   time   to   charge :   q cap
= q
0
(
1 − e −
)
• Where   e is   Euler’s   constant   ( e =2.71828…)
• q
0
= CV
0
,   where   V battery.
0 is   the   voltage   of   the
• The   capacitor   will   reach   full   charge   in   infinite   time   (although   in   reality   we   can   set   a   percentage   of   full   charge   to   get   an   acceptable   finite   time).
Charging   and   Discharging
Capacitors   in   a   Circuit
• If   a   Direct   Current   circuit   contains   a   resistor,    an   initially   charged capacitor   and   a   switch,   then   the   capacitor   will   take   time   to   discharge :   q cap
= q e −
• Where   e is   Euler’s   constant   ( e =2.71828…)
• q
0   is   the   initial   charge   on   the   battery.
• The   capacitor   will   discharge   in   infinite   time
(although   in   reality   we   can   set   a   low   percentage   of   full   charge   to   get   an   acceptable   finite   time).
3/17/2009
29

Charging   and   Discharging
Capacitors   in   a   Circuit
• Charging   Capacitor q cap
= q
0
(
1 − e −
)
• Discharging   Capacitor q cap
= q e −
• τ =RC ,   is   the   time   constant of   the   circuit(s).
• When   RC is   small,   the   capacitor   charges/discharges   quickly
• When   RC is   large,   the   capacitor   will   charge/discharge   slowly.
Examples
20.97
– The   circuit   in   the   drawing   contains   two   resistors   and two   capacitors   that   are   connected   to   a   battery   via   a   switch.
When   the   switch   is   closed,   the   capacitors   begin   to   charge up.
What   is   the   time   constant   for   the   charging   process?
Find   the   equivalent   resistance   and   capacitance:
C
S
= C
1
+ C
2
1
=
1
+
1
R
P
R
1
R
2
C
S
= μ
C
S
=
μ
1
R
P
R p
=
4.0
1 k Ω
= 1.333
k Ω
+
1
2.0
k Ω
The   time   constant   is   the   product   of   equivalent   resistance   and   equivalent   capacitance
τ = R C
S
=
(
× 3 Ω
)(
× −
)
τ = 1.2 10 s
3/17/2009
30

Examples
20.98
– How   many   time   constants   must   elapse   before   a   capacitor   in   a   series   RC   circuit   is   charged   to   80.0%   of   its   equilibrium   charge?
Use   the   Charging   equation: q cap
= q
0
(
1 − e −
)
We   want   q cap to   be 0.8
q
0
,
0.8
q
0
= q
0
(
1 − e − t / τ
)
0 .2
= e − t / τ t = −
⇒
t
( )
= 1.6
1
3/17/2009
31