1
00:00:00,000 --> 00:00:01,600


2
00:00:01,600 --> 00:00:04,210
Having to deal with a
single capacitor hooked up

3
00:00:04,210 --> 00:00:06,830
to a battery isn't
all that difficult,

4
00:00:06,830 --> 00:00:09,180
but when you have
multiple capacitors,

5
00:00:09,180 --> 00:00:12,800
people typically get
much, much more confused.

6
00:00:12,800 --> 00:00:14,550
There's all kinds
of different ways

7
00:00:14,550 --> 00:00:16,540
to hook up multiple capacitors.

8
00:00:16,540 --> 00:00:18,630
But if capacitors
are connected one

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00:00:18,630 --> 00:00:20,850
after the other in
this way, we call

10
00:00:20,850 --> 00:00:23,870
them capacitors
hooked up in series.

11
00:00:23,870 --> 00:00:26,400
So say you were taking
a test, and on the test

12
00:00:26,400 --> 00:00:30,600
it asked you to find the charge
on the leftmost capacitor.

13
00:00:30,600 --> 00:00:33,130
What some people might
try to do is this.

14
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Since capacitance is the
charge divided by the voltage,

15
00:00:36,750 --> 00:00:38,580
they might plug
in the capacitance

16
00:00:38,580 --> 00:00:41,895
of the leftmost capacitor,
which is 4 farads, plug

17
00:00:41,895 --> 00:00:44,960
in the voltage of the
battery, which is 9 volts.

18
00:00:44,960 --> 00:00:48,310
Solving for the charge, they'd
get that the leftmost capacitor

19
00:00:48,310 --> 00:00:53,550
stores 36 coulombs, which
is totally the wrong answer.

20
00:00:53,550 --> 00:00:56,110
To try and figure out
why and to figure out

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00:00:56,110 --> 00:00:59,280
how to properly deal with
this type of scenario,

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00:00:59,280 --> 00:01:02,560
let's look at what's actually
going on in this example.

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When the battery's hooked
up, a negative charge

24
00:01:05,040 --> 00:01:08,210
will start to flow from
the right side of capacitor

25
00:01:08,210 --> 00:01:11,400
3, which makes a negative
charge get deposited

26
00:01:11,400 --> 00:01:13,920
on the left side of capacitor 1.

27
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This makes a
negative charge flow

28
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from the right
side of capacitor 1

29
00:01:18,120 --> 00:01:20,800
on to the left side
of capacitor 2.

30
00:01:20,800 --> 00:01:23,040
And that makes a
negative charge flow

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from the right
side of capacitor 2

32
00:01:25,066 --> 00:01:28,030
on to the left side
of capacitor 3.

33
00:01:28,030 --> 00:01:30,200
Charges will
continue doing this.

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And it's important to
note something here.

35
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Because of the way the
charging process works,

36
00:01:35,270 --> 00:01:38,460
all of the capacitors here
must have the same amount

37
00:01:38,460 --> 00:01:40,180
of charge stored on them.

38
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It's got to be that way.

39
00:01:41,850 --> 00:01:44,090
Looking at how these
capacitors charge up,

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there's just nowhere
else for the charge

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00:01:46,260 --> 00:01:49,240
to go but on to the next
capacitor in the line.

42
00:01:49,240 --> 00:01:50,950
This is actually good news.

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00:01:50,950 --> 00:01:54,010
This means that for
capacitors in series,

44
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the charge stored
on every capacitor

45
00:01:56,400 --> 00:01:58,030
is going to be the same.

46
00:01:58,030 --> 00:02:00,810
So if you find the charge
on one of the capacitors,

47
00:02:00,810 --> 00:02:03,970
you've found the charge
on all of the capacitors.

48
00:02:03,970 --> 00:02:06,540
But how do we figure out
what that amount of charge

49
00:02:06,540 --> 00:02:07,640
is going to be?

50
00:02:07,640 --> 00:02:09,550
Well, there's a
trick we can use when

51
00:02:09,550 --> 00:02:12,040
dealing with
situations like this.

52
00:02:12,040 --> 00:02:15,240
We can imagine replacing
our three capacitors

53
00:02:15,240 --> 00:02:18,250
with just a single
equivalent capacitor.

54
00:02:18,250 --> 00:02:21,660
If we choose the right value
for this single capacitor,

55
00:02:21,660 --> 00:02:24,060
then it will store the
same amount of charge

56
00:02:24,060 --> 00:02:27,340
as each of the three
capacitors in series will.

57
00:02:27,340 --> 00:02:29,430
The reason this is
useful is because we

58
00:02:29,430 --> 00:02:32,230
know how to deal with
a single capacitor.

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00:02:32,230 --> 00:02:34,880
We call this imaginary
single capacitor

60
00:02:34,880 --> 00:02:37,530
that's replacing
multiple capacitors

61
00:02:37,530 --> 00:02:39,690
the "equivalent capacitor."

62
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It's called the
equivalent capacitor

63
00:02:41,640 --> 00:02:43,520
because its effect
on the circuit

64
00:02:43,520 --> 00:02:47,240
is, well, equivalent
to the sum total effect

65
00:02:47,240 --> 00:02:50,400
that the individual capacitors
have on the circuit.

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00:02:50,400 --> 00:02:52,710
And it turns out that
there's a handy formula that

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lets you determine the
equivalent capacitance.

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The formula to find the
equivalent capacitance

69
00:02:58,040 --> 00:03:01,710
of capacitors hooked up
in series looks like this.

70
00:03:01,710 --> 00:03:03,990
1 over the equivalent
capacitance

71
00:03:03,990 --> 00:03:07,210
is going to equal 1 over
the first capacitance

72
00:03:07,210 --> 00:03:10,430
plus 1 over the second
capacitance plus 1

73
00:03:10,430 --> 00:03:12,170
over the third capacitance.

74
00:03:12,170 --> 00:03:15,640
And if you had more capacitors
that were in that same series,

75
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you would just
continue on this way

76
00:03:17,280 --> 00:03:19,730
until you've included
all of the contributions

77
00:03:19,730 --> 00:03:21,340
from all of the capacitors.

78
00:03:21,340 --> 00:03:24,480
We'll prove where this formula
comes from in a minute,

79
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but for now, let's just
get used to using it

80
00:03:26,670 --> 00:03:28,610
and see what we can figure out.

81
00:03:28,610 --> 00:03:30,600
Using the values
from our example,

82
00:03:30,600 --> 00:03:33,400
we get that 1 over the
equivalent capacitance

83
00:03:33,400 --> 00:03:37,760
is going to be 1 over 4
farads plus 1 over 12 farads

84
00:03:37,760 --> 00:03:42,100
plus 1 over 6 farads,
which equals 0.5.

85
00:03:42,100 --> 00:03:43,060
But be careful.

86
00:03:43,060 --> 00:03:44,170
You're not done yet.

87
00:03:44,170 --> 00:03:46,640
We want the equivalent
capacitance, not

88
00:03:46,640 --> 00:03:48,840
1 over the equivalent
capacitance.

89
00:03:48,840 --> 00:03:53,470
So we have to take 1 over this
value of 0.5 that we found.

90
00:03:53,470 --> 00:03:56,330
And if we do that, we get that
the equivalent capacitance

91
00:03:56,330 --> 00:03:59,980
for this series of
capacitors is 2 farads.

92
00:03:59,980 --> 00:04:03,270
Now that we've reduced our
complicated multiple capacitor

93
00:04:03,270 --> 00:04:06,060
problem into a single
capacitor problem,

94
00:04:06,060 --> 00:04:08,920
we can solve for the charge
stored on this equivalent

95
00:04:08,920 --> 00:04:09,840
capacitor.

96
00:04:09,840 --> 00:04:12,910
We can use the formula
capacitance equals charge

97
00:04:12,910 --> 00:04:16,180
per voltage and plug in
the value of the equivalent

98
00:04:16,180 --> 00:04:17,110
capacitance.

99
00:04:17,110 --> 00:04:19,470
And we can plug in the
voltage of the battery

100
00:04:19,470 --> 00:04:23,110
now because the voltage across
a single charged-up capacitor

101
00:04:23,110 --> 00:04:25,980
is going to be the same as the
voltage of the battery that

102
00:04:25,980 --> 00:04:27,060
charged it up.

103
00:04:27,060 --> 00:04:29,610
Solving for the charge, we
get that the charge stored

104
00:04:29,610 --> 00:04:33,120
on this equivalent
capacitor is 18 coulombs.

105
00:04:33,120 --> 00:04:35,950
But we weren't trying to find
the charge on the equivalent

106
00:04:35,950 --> 00:04:36,660
capacitor.

107
00:04:36,660 --> 00:04:38,220
We were trying to
find the charge

108
00:04:38,220 --> 00:04:40,050
on the leftmost capacitor.

109
00:04:40,050 --> 00:04:42,090
But that's easy now
because the charge

110
00:04:42,090 --> 00:04:45,630
on each of the individual
capacitors in series

111
00:04:45,630 --> 00:04:48,140
is going to be the same as
the charge on the equivalent

112
00:04:48,140 --> 00:04:48,990
capacitor.

113
00:04:48,990 --> 00:04:51,340
So since the charge on
the equivalent capacitor

114
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was 18 coulombs,
the charge on each

115
00:04:54,180 --> 00:04:56,640
of the individual
capacitors in series

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00:04:56,640 --> 00:04:58,930
is going to be 18 coulombs.

117
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This process can be
confusing to people,

118
00:05:01,200 --> 00:05:03,210
so let's try another example.

119
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This time, let's say you had
four capacitors hooked up

120
00:05:06,260 --> 00:05:08,830
in series to a 24-volt battery.

121
00:05:08,830 --> 00:05:10,990
The arrangement of
these capacitors

122
00:05:10,990 --> 00:05:13,490
looks a little different
from the last example,

123
00:05:13,490 --> 00:05:16,200
but all of these capacitors
are still in series

124
00:05:16,200 --> 00:05:19,020
because they're hooked up
one right after the other.

125
00:05:19,020 --> 00:05:21,460
In other words, the
charge has no choice

126
00:05:21,460 --> 00:05:24,170
but to flow directly
from one capacitor

127
00:05:24,170 --> 00:05:26,190
straight to the next capacitor.

128
00:05:26,190 --> 00:05:30,270
So these capacitors are still
considered to be in series.

129
00:05:30,270 --> 00:05:32,320
Let's try to figure
out the charge that's

130
00:05:32,320 --> 00:05:35,980
going to be stored on
the 16-farad capacitor.

131
00:05:35,980 --> 00:05:38,470
We'll use the same
process as before.

132
00:05:38,470 --> 00:05:41,330
First we imagine replacing
the four capacitors

133
00:05:41,330 --> 00:05:43,590
with a single
equivalent capacitor.

134
00:05:43,590 --> 00:05:45,880
We'll use the formula
to find the equivalent

135
00:05:45,880 --> 00:05:48,480
capacitance of
capacitors in series.

136
00:05:48,480 --> 00:05:51,720
Plugging in our values, we
find that 1 over the equivalent

137
00:05:51,720 --> 00:05:55,860
capacitance is going
to equal 0.125.

138
00:05:55,860 --> 00:05:56,440
Be careful.

139
00:05:56,440 --> 00:05:58,800
We still have to take
1 over this value

140
00:05:58,800 --> 00:06:01,630
to get that the equivalent
capacitance for this circuit

141
00:06:01,630 --> 00:06:03,747
is going to be 8 farads.

142
00:06:03,747 --> 00:06:05,580
Now that we know the
equivalent capacitance,

143
00:06:05,580 --> 00:06:07,730
we can use the
formula capacitance

144
00:06:07,730 --> 00:06:09,610
equals charge per voltage.

145
00:06:09,610 --> 00:06:12,940
We can plug in the value of
the equivalent capacitance, 8

146
00:06:12,940 --> 00:06:13,500
farads.

147
00:06:13,500 --> 00:06:15,500
And since we have a
single capacitor now,

148
00:06:15,500 --> 00:06:17,370
the voltage across
that capacitor

149
00:06:17,370 --> 00:06:20,340
is going to be the same as the
voltage of the battery, which

150
00:06:20,340 --> 00:06:21,690
is 24 volts.

151
00:06:21,690 --> 00:06:24,620
So we find that our imaginary
equivalent capacitor

152
00:06:24,620 --> 00:06:28,550
would store a charge
of 192 coulombs.

153
00:06:28,550 --> 00:06:30,290
This means that
the charge on each

154
00:06:30,290 --> 00:06:35,270
of the individual capacitors is
also going to be 192 coulombs.

155
00:06:35,270 --> 00:06:37,260
And this gives us our
answer, that the charge

156
00:06:37,260 --> 00:06:42,110
on the 16-farad capacitor
is going to be 192 coulombs.

157
00:06:42,110 --> 00:06:43,880
In fact, we can go even further.

158
00:06:43,880 --> 00:06:46,290
Now that we know the
charge on each capacitor,

159
00:06:46,290 --> 00:06:48,000
we can solve for
the voltage that's

160
00:06:48,000 --> 00:06:51,600
going to exist across each
of the individual capacitors.

161
00:06:51,600 --> 00:06:53,940
We'll again use the
fact that capacitance

162
00:06:53,940 --> 00:06:55,980
is the charge per voltage.

163
00:06:55,980 --> 00:06:58,760
If we plug in the values
for capacitor one,

164
00:06:58,760 --> 00:07:02,060
we'll plug in a
capacitance of 32 farads.

165
00:07:02,060 --> 00:07:06,260
The charge that capacitor
one stores is 192 coulombs.

166
00:07:06,260 --> 00:07:09,520
So we can solve for the
voltage across capacitor 1,

167
00:07:09,520 --> 00:07:11,660
and we get 6 volts.

168
00:07:11,660 --> 00:07:14,110
If we were to do the
same calculation for each

169
00:07:14,110 --> 00:07:17,860
of the other three capacitors,
always being careful that we

170
00:07:17,860 --> 00:07:20,410
use their particular
values, we'll

171
00:07:20,410 --> 00:07:22,770
get that the voltages
across the capacitors

172
00:07:22,770 --> 00:07:26,840
are 2 volts across the
96-farad capacitor,

173
00:07:26,840 --> 00:07:30,240
12 volts across the
16-fard capacitor,

174
00:07:30,240 --> 00:07:33,730
and 4 volts across the
48-farad capacitor.

175
00:07:33,730 --> 00:07:35,670
Now, the real reason I
had us go through this

176
00:07:35,670 --> 00:07:37,750
is because I wanted to
show you something neat.

177
00:07:37,750 --> 00:07:40,630
If you add up the voltages
that exist across each

178
00:07:40,630 --> 00:07:43,960
of the capacitors,
you'll get 24 volts,

179
00:07:43,960 --> 00:07:47,610
the same as the
value of the battery.

180
00:07:47,610 --> 00:07:49,070
This is no coincidence.

181
00:07:49,070 --> 00:07:52,020
If you add up the voltages
across the components

182
00:07:52,020 --> 00:07:55,910
in any single-loop circuit like
this, the sum of the voltages

183
00:07:55,910 --> 00:07:59,260
is always going to equal
the voltage of the battery.

184
00:07:59,260 --> 00:08:01,100
And this principle
will actually let

185
00:08:01,100 --> 00:08:04,280
us derive the formula we've
been using for the equivalent

186
00:08:04,280 --> 00:08:07,020
capacitance of
series capacitors.

187
00:08:07,020 --> 00:08:08,700
To derive this
formula, let's say

188
00:08:08,700 --> 00:08:14,770
we've got three capacitors with
capacitances of C1, C2, and C3

189
00:08:14,770 --> 00:08:18,440
hooked up in series to a
battery of voltage V. We now

190
00:08:18,440 --> 00:08:21,860
know that if we add up the
voltage across each capacitor,

191
00:08:21,860 --> 00:08:24,460
it's got to add up to the
voltage of the battery.

192
00:08:24,460 --> 00:08:26,600
Using the formula
for capacitance,

193
00:08:26,600 --> 00:08:30,290
we can see that the voltage
across an individual capacitor

194
00:08:30,290 --> 00:08:33,130
is going to be the charge
on that capacitor divided

195
00:08:33,130 --> 00:08:34,440
by its capacitance.

196
00:08:34,440 --> 00:08:36,780
So the voltage
across each capacitor

197
00:08:36,780 --> 00:08:41,760
is going to be Q over C1,
Q over C2, and Q over C3,

198
00:08:41,760 --> 00:08:42,740
respectively.

199
00:08:42,740 --> 00:08:46,140
I didn't write Q1, Q2,
or Q3 because remember,

200
00:08:46,140 --> 00:08:48,810
all the charges on
capacitors in series

201
00:08:48,810 --> 00:08:50,270
are going to be the same.

202
00:08:50,270 --> 00:08:53,630
These voltages have to add up
to the voltage of the battery.

203
00:08:53,630 --> 00:08:55,930
I can pull out a
common factor of Q

204
00:08:55,930 --> 00:08:58,210
because it's in each
term on the left.

205
00:08:58,210 --> 00:09:00,700
And now I'm going to
divide each side by Q.

206
00:09:00,700 --> 00:09:02,260
I did that because
look at what we've

207
00:09:02,260 --> 00:09:04,690
got on the right-hand
side of this equation.

208
00:09:04,690 --> 00:09:08,370
The voltage across the battery
divided by the charge stored

209
00:09:08,370 --> 00:09:12,500
is just equal to 1 over
the equivalent capacitance,

210
00:09:12,500 --> 00:09:16,570
because Q over V is equal to
the equivalent capacitance.

211
00:09:16,570 --> 00:09:17,410
And here it is.

212
00:09:17,410 --> 00:09:19,210
This is the formula
we've been using,

213
00:09:19,210 --> 00:09:20,820
and this is where it comes from.

214
00:09:20,820 --> 00:09:23,290
It's derived from the
fact that the voltages

215
00:09:23,290 --> 00:09:25,490
across these
capacitors in series

216
00:09:25,490 --> 00:09:29,020
have to add up to the
voltage of the battery.

217
00:09:29,020 --> 00:00:00,000