Capacitor
•  Parallel plates, separated by an insulator, so no charge
flows between the plates. Impose a time-varying voltage
drop:
→ time-varying electric field
→ time-varying displacement current
•  Capacitor equation:
dv (t )
i (t ) = C
dt
•  Units: v(t) is volts, i(t) is amps, and C is farads [F]
Look at the capacitor equation again:
dv (t )
i (t ) = C
dt
Suppose v(t) is constant. Then i(t) =
A.  0
B.  ∞
C.  a constant
So, if the voltage drop across the capacitor
is constant, its current is 0, so the
capacitor can be replaced by
A.  a short circuit
B.  an open circuit
C.  a constant
Capacitor
•  If the voltage drop across a capacitor is constant,
the current is 0, so the capacitor can be replaced
by an OPEN CIRCUIT.
dv (t )
•  Look at the capacitor equation again: i (t ) = C
dt
•  Suppose there is a discontinuity in v(t) – that is, at some
value of t, the voltage jumps instantaneously. At this
value of t, the derivative of the voltage is infinite.
Therefore the current is infinite! NOT POSSIBLE.
•  Thus, the voltage drop across a capacitor is
continuous for all time.
The voltage drop across a capacitor is
described as
v(t ) = Vo , t < 0
v(t ) = 15e −250t − 10e −1000t V, t ≥ 0
This expression for the capacitor voltage is
valid only if the value of Vo is
A.  0 V
B.  25 V
C.  5 V
Capacitor
•  The equation for voltage in terms of current:
dv (t )
i (t ) = C
dt
i (t )dt = Cdv(t )
⇒
t
v(t )
to
v ( to )
⇒
∫ i(τ )dτ = C ∫
dx
⇒
1 t
v (t ) = ∫ i (τ )dτ + v (to )
C to
Capacitor
•  Power and energy
dv (t )
p(t ) = v (t )i (t ) = Cv (t )
dt
dw(t )
dv (t )
p(t ) =
= Cv (t )
dt
dt
⇒
dw(τ ) = Cv (τ )dv (τ )
w( t )
dx = C ∫
v(t )
⇒
∫
⇒
w(t ) = 12 Cv (t )2
0
0
y (τ )dτ
(passive sign convention! )
Capacitor
Ø Capacitors, like inductors, are energy storage
devices!
•  If the initial voltage drop across the capacitor is nonzero, the capacitor is storing energy.
Suppose the initial voltage drop across a 4
µF capacitor is 10 V. The initial energy
stored in the capacitor is
A.  400 µJ
B.  200 µJ
C.  20 µJ
Capacitor
•  Example 6.5 – find the voltage, power, and energy
For t < 0 :
v (t ) = 0 V;
p(t ) = 0 W;
w(t ) = 0 J
For 0 ≤ t ≤ 20 µs :
t
v (t ) =
1
1 5000 x
5000
x
dx
+
v
(
0
)
=
0.2 µ ∫0
0.2 µ
2
2 t
= 12.5 × 109 t 2 V
0
p(t ) = v (t )i (t ) = (12.5 × 109 t 2 )(5000t ) = 62.5 × 1012 t 3 W
w(t ) = 12 (0.2 µ )(12.5 × 109 t 2 )2 = 15.625 × 1012 t 4 J
At t = 20 µs,
v ( 20 µs) = 12.5 × 109 ( 20 µ)2 = 5 V
Capacitor
•  Example 6.5, continued
For 20 µs ≤ t ≤ 40 µs :
t
1
v (t ) =
(0.2 − 5000 x ) dx + v ( 20 µ )
∫
0.2 µ 20 µ
2
=
t
1 ⎡
5000 x ⎤
0
.
2
x
−
+5
⎢
⎥
0.2 µ ⎣
2 ⎦ 0
= (10 6 t − 12.5 × 10 9 t 2 − 10) V
p(t ) = v (t )i (t );
At t = 40 µs,
w(t ) = 12 Cv (t ) 2
v ( 40 µs) = [106 ( 40 µ ) − 12.5 × 109 ( 40 µ ) 2 − 10) = 10 V
Capacitor
•  Example 6.5, continued
For t ≥ 40 µs :
t
1
v (t ) =
0 dx + v(40 µ ) = 10 V
∫
0.2 µ 40 µ
p(t ) = v(t )i (t ) = 0 W
w(t ) = 12 (0.2 µ )(10)2 = 10 µJ!
During the interval between 0 and 40µs, the power is positive
(absorbed), energy is stored and “trapped” by the capacitor, so
even when the current goes to 0, the voltage stays at 10 V and the
energy is non-zero.
Capacitor
•  Example 6.5, continued
Capacitors in series and parallel
KCL :
dv
dv
dv
+ C2
+ C3
dt
dt
dt
dv
dv
= (C1 + C2 + C3 ) = Ceq
dt
dt
i = i1 + i2 + i3 = C1
Capacitors in parallel ADD.
Capacitors in series and parallel
1 t
KVL : v = v1 + v2 + v3 = ∫ i (τ )dτ + v1 (to )
C1 to
1
+
C2
t
∫ i(τ )dτ + v (t )
2
to
o
1 t
+ ∫ i (τ )dτ + v3 (to )
C3 to
⎛ 1
1
1 ⎞ t
= ⎜⎜ +
+ ⎟⎟ ∫ i (τ )dτ
⎝ C1 C2 C3 ⎠ to
+ [v1 (to ) + v2 (to ) + v3 (to )]
1
=
Ceq
t
∫ i(τ )dτ + v
to
eq
(to )
Capacitors combine in series like resistors or inductors
combine in parallel.
Find the equivalent capacitance for the
circuit below, assuming v1 = 12 V and v2 = -8
V.
A.  Ceq = 4µF, veq = 4 V
B.  Ceq = 9.5µF, veq = 4 V
C.  Ceq = 4µF, veq = 20 V
Inductor and Capacitor comparison
Inductor
Capacitor
Henries [H]
Farads [F]
Symbol
Units
Describing
equation
Other
equation
v (t ) = L
di (t )
dt
i (t ) = C
dv (t )
dt
1 t
1 t
i (t ) = ∫ v(τ )dτ + i (to ) v(t ) = ∫ i (τ )dτ + v(to )
L to
C to
Initial
condition
i(to)
v(to)
Behavior with
const. source
If i(t) = I, v(t) = 0
→ short circuit
If v(t) = V, i(t) = 0
→ open circuit
Continuity
requirement
i(t) is continuous so v(t)
is finite
v(t) is continuous so i(t)
is finite
Inductor and Capacitor comparison
Inductor
Capacitor
Power
di (t )
dv (t )
p(t ) = v(t )i (t ) = Li (t )
p(t ) = v(t )i (t ) = Cv(t )
dt
dt
Energy
w(t ) = 12 Li (t ) 2
w(t ) = 12 Cv (t ) 2
Initial
energy
wo (t ) = 12 Li (to ) 2
wo (t ) = 12 Cv (to ) 2
Trapped
energy
w(∞) = 12 Li (∞)2
w( ∞ ) = 12 Cv ( ∞) 2
Seriesconnecte
d
Parallelconnecte
d
Leq = L1 + L2 + L2
1
1 1 1
= + +
Ceq C1 C2 C3
ieq (to ) = i (to )
veq (to ) = v1 (to ) + v2 (to ) + v3 (to )
1
1 1 1
= + +
Leq L1 L2 L3
Ceq = C1 + C2 + C2
ieq (to ) = i1 (to ) + i2 (to ) + i3 (to )
veq (to ) = v(to )
Chapter 6 – Inductance and Capacitance
•  Objectives:
•  Inductor
Ø Equations for v, i, p, and w
Ø Behavior in the presence of constant current
Ø Continuity requirements
Ø Combine in series and in parallel
•  Capacitor
Ø Equations for v, i, p, and w
Ø Behavior in the presence of constant voltage
Ø Continuity requirements
Ø Combine in series and in parallel
Inductor
•  Coil of wire with time-varying current
→ time-varying magnetic field
→ time-varying voltage drop
•  Inductor equation:
di (t )
v (t ) = L
dt
•  Units: v(t) is volts, i(t) is amps, and L is henries [H]
Look at the inductor equation again:
di (t )
v (t ) = L
dt
Suppose i(t) is constant. Then v(t) =
A.  L
B.  0
C.  Undefined
So, if the current in an inductor is constant,
its voltage drop is 0, so the inductor can be
replaced by
A.  A short circuit
B.  An open circuit
C.  A capacitor
Inductor
•  If the current in the inductor is constant, the voltage
is 0, so the inductor can be replaced by a SHORT
CIRCUIT.
•  Look at the inductor equation again:
di (t )
v (t ) = L
dt
•  Suppose there is a discontinuity in i(t) – that is, at some value
of t, the current jumps instantaneously. At this value of t, the
derivative of the current is infinite. Therefore the voltage is
infinite! NOT POSSIBLE.
•  Thus, the current through an inductor is continuous
for all time.
We just showed that the current in an
inductor must be continuous for all time.
This means that the voltage must also be
continuous for all time.
A.  True
B.  False
Inductor
•  The equation for current in terms of voltage:
di (t )
v (t ) = L
dt
v (t )dt = Ldi(t )
⇒
t
i(t )
to
i ( to )
⇒
∫ v(τ )dτ = L ∫
dx
⇒
1 t
i (t ) = ∫ v (τ )dτ + i (to )
L to
Inductor
•  Power and energy
di (t )
p(t ) = v (t )i (t ) = Li (t )
dt
dw(t )
di (t )
p(t ) =
= Li (t )
dt
dt
⇒
dw(τ ) = Li (τ )di (τ )
w( t )
dx = L ∫
i(t )
⇒
∫
⇒
w(t ) = 12 Li (t )2
0
0
y (τ )dτ
(passive sign convention! )
Inductor
Ø Inductors are energy storage devices!
•  If the initial current through the inductor is non-zero,
the inductor is storing energy.
Suppose the initial current through a 2 mH
inductor is 1 A. The initial energy stored in
the inductor is
A.  1 mJ
B.  2 mJ
C.  4 mJ
Inductor
•  Example – Assessment Problem 6.1
Is the current continuous? i(0) = 8 – 8 = 0: YES!
Find the voltage:
v (t ) = 0,
t<0
di (t )
v (t ) = L
= (0.004)[( −300)8e −300t − ( −1200)8e −1200 t ]
dt
= −9.6e −300t + 38.4e −1200 t V,
t≥0
Is the voltage continuous? v(0) = -9.6 + 38.4 = 28.8 V:
NO!
Inductor
•  Example – Assessment Problem 6.1, continued
Find the power for the inductor:
p(t ) = v (t )i (t )
= ( −9.6e −300t + 38.4e −1200 t )(8e −300t − 8e −1200 t )
= −76.8e −600t + 384e −1500 t − 307.2e −2400 t W
To find the max power and the time at which the power
is max, take the first derivative of p(t) and set it equal
to 0.
Inductor
•  Example – Assessment Problem 6.1, continued
Find the energy for the inductor:
w(t ) = 12 Li (t )2
= 12 (0.004)(8e −300t − 8e −1200t )2
= 128(e −600t − 2e −1500t + e −2400t ) mJ
To find the max energy and the time at which the
energy is max, take the first derivative of w(t) and set it
equal to 0. Or if you don’t have w(t) yet, do the same
for i(t)!
Inductors in series and parallel
KVL:
Inductors in series ADD.
Inductors in series and parallel
KCL:
Inductors combine in parallel like
resistors combine in parallel.
Find the equivalent inductance for the
circuit below, assuming i1= 6 A and i2 = -3 A.
A.  Leq = 300 mH, i(t) = 6 A
B.  Leq = 270 mH, i(t) = -3 A
C.  Leq = 270 mH, i(t) = 3 A
Inductor summary
Symbol
Units
Describing equation
Henries [H]
v (t ) = L
di (t )
dt
Other equation
1 t
i (t ) = ∫ v(τ )dτ + i (to )
L to
Initial condition
i(to)
Behavior with const.
source
If i(t) = I, v(t) = 0 → short circuit
Continuity requirement
i(t) is continuous so v(t) is finite
Inductor summary
Power
Energy
p(t ) = v(t )i (t ) = Li (t )
w(t ) = 12 Li (t ) 2
Initial energy
wo (t ) = 12 Li (to ) 2
Trapped energy
w(∞) = 12 Li (∞)2
Series-connected
Parallel-connected
di (t )
dt
Leq = L1 + L2 + L2
ieq (to ) = i (to )
1
1 1 1
= + +
Leq L1 L2 L3
ieq (to ) = i1 (to ) + i2 (to ) + i3 (to )