MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Fall 2007
Problem Set 11 Solutions
Problem 1: Where’s the energy go?
Consider a simple RC circuit consisting of a “variable” battery (EMF 0 V or
ε
), a resistor
R
and an uncharged capacitor
C
all in series.
At time t=0 the battery is “turned on” (EMF
ε
) and the capacitor is allowed to charge.
A time
T
(>>
RC
) later the battery is turned off
(EMF 0 V) and the capacitor discharges.
This process is continually repeated.
(a) How much energy does the resistor dissipate while the capacitor is being charged (i.e.
between t=0 and t=
T
)?
The resistor dissipates power
P
=
I
2
R
. To find the energy dissipated we need to integrate:
() ()
N
2
2
00
0
22
2
2
2
1
2
0
0
0
1
TT
T
t
tt
t
T
T
t
tT
t
U
P t dt
I
t Rdt
e
Rdt
R
e
ed
t
e
C
RR
R
τ
ττ
ε
εε
−
==
=
−
−−
≈
=
⎛⎞
Δ=
=
=
⎜⎟
⎝⎠
−
=
−
≈
∫∫
∫
∫
(b) How does this compare to the amount of energy stored in the capacitor at time
T
?
They are equal.
(c) How much energy does the resistor dissipate while the capacitor is discharging (i.e.
between t=
T
and t=2
T
)?
Since it gets rid of all the energy in the capacitor during this time, it’s the same:
2
1
2
C
(d) What is the time average power dissipated by the resistor?
Hint:
To average a set of
numbers you add the numbers and divide by the number you added.
To average a
periodic function, integrate it over one period and divide by the length (or in this case
time
) of the period.
We already did the integral.
It dissipated
2
1
2
C
every T so
2
1
2
T
PC
=
Problem 2:
Capacitor discharge
You are tasked with designing a capacitor (capacitance
C
) that will hold its charge for a
long time.
You want to make it using a dielectric with dielectric constant
κ
and
resistivity
ρ
.
That resistivity is a problem – even though it is high, it still allows charge to
flow from one plate to the other over time.
So you want to maximize the resistance
between the plates.
PS11 Solutions1
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View Full DocumentProblem 2:
Capacitor discharge
continued
…
(a)
Compare the resistance of the dielectric between the plates of a parallel plate,
cylindrical and spherical capacitor, each with capacitance C but with whatever
dimensions you want.
HINT:
Calculate
RC
and solve for
R
.
We have previously calculated the capacitance for each of these geometries, but I’ll
briefly remind you of the steps.
Parallel Plate Capacitor (area
A
, separation
d
)
00
0
0
;
;
( &
don't matter)
A
QQ
d
d
EV
C
R
R
C
R
A
d
AA
d
A
C
κε
ρκε
σρ
==
⇒
=
⇒
=
=
=
⇒
=
Cylindrical Capacitor (inner radius
a
, outer radius
b
, length
L
)
()
0
2
ln
22
l
n
L
b
C
rL
L
a
b a
πκε
⎛⎞
=⇒
=
⇒
=
⎜⎟
⎝⎠
You already calculated the resistance in this situation in pset #6, problem 3, but as a
reminder, to calculate the resistance we consider the dielectric to consist of a number of
cylindrical shells, all in series with each other.
Consider a thin cylindrical shell of radius
r
, thickness
dr
, and length
L
. Its contribution to the overall resistance is
dd
r
dR
d
r
A
rL
L r
ρρ
ρ
ππ
=
A
The resistance of the whole annulus is the series summation of the contributions of the
thin shells:
0
0
ln
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 Fall '07
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