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reproduction or display.
  the
ideal capacitor is a passive element with
circuit symbol
  the
current-voltage relation is
  the
capacitance C is measured in farads (F)
dv
i =C
dt
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2
  capacitors
can be bulky and typical values
range from pF to µF
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3
Since
⎛ dv ⎞
dw
p(t) = i(t)v(t) = ⎜C
⎟v =
⎝ dt ⎠
dt
then the energy stored in a capacitor is
w = Cv
1
2
2
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  capacitors
  the
are open circuits to dc voltages
voltage on a capacitor cannot jump
  capacitors
store energy (iv>0) or deliver energy
(iv<0)
dv
i =C
dt
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5
Find i(t) for the voltages
shown, if C=2 F.
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Solution: apply i(t)=2dv/dt and graph:
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7
Show that the following graphs are matching
voltage and current graphs for a capacitor of
C=5µF.
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8
Determine the maximum energy stored in the
capacitor, and plot iR and iC.
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  the
ideal inductor is a passive element with
circuit symbol
  the
current-voltage relation is
di
v=L
dt
  the
unit of inductance L is henry (H)
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  inductors
can be bulky and typical values
range from µH to H
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Since
⎛ di ⎞
dw
p(t) = i(t)v(t) = ⎜ L ⎟i =
⎝ dt ⎠
dt
then the energy stored in a inductor is
w = Li
1
2
2
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  inductors
  the
are short circuits to dc voltages
current through an inductor cannot jump
  inductors
store energy (iv>0) or deliver energy
(iv<0)
di
v=L
dt
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13
Show that the following graphs are matching
voltage and current graphs for an inductor of
L=3 H.
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14
For the same 3-H inductor, the voltages are 10
times larger when the current is ramped 10
times faster:
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15
Determine the maximum energy stored in the
inductor, and find the energy lost to resistor
from t=0 to t=6 s.
Answer: 216 J, 43.2 J
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16
Apply KVL to show:
Leq = L1 + L2 + + LN
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Apply KCL to show
Leq =
1
1
1 + + 1
+
L1
L2
LN
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18
Apply KVL to show:
Ceq =
1
1
1 + + 1
+
C1
C2
CN
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Apply KCL to show:
Ceq = C1 + C2 + + CN
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Two capacitors in series:
C1C2
Ceq =
C1 + C2
Two inductors in parallel:
L1L2
Leq =
L1 + L2
Two resistors in parallel:
R1R2
Req =
R1 + R2
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21
Show that these circuits are
equivalent using series and
parallel combinations.
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vout
1
=−
R1 C f
!
t
"
vs dt − vC f (0)
0
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23
vout
dvs
= −C1 R f
dt
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