186 Week 5: Resistance
Multiply through bydt, divide through byQ, to get:
dQ
Q
=−dt
RC
(376)
Integrate both sides (indefinite integral on the right):
ln(Q) =−
t
RC
+A (377)
(whereAis the constant of integration). To getQ, we exponentiate both sides:
Q(t) =eln(Q)=e−
RCt+A
=eAe−t/RC (378)
Finally, we set the constant of integration from the initial conditions, so thatQ(0) =Q 0 :
Q(t) =Q 0 e−t/RC (379)
From this we can easily find the other quantities mentioned above:
I(t) = −
dQ
dt
=
Q 0
RC
e−t/RC (380)
VC(t) = Q
C
=Q^0
C
e−t/RC=V 0 e−t/RC (381)
VR(t) = −I(t)R=−
Q 0
C
e−t/RC (382)
PC(t) = VC(t)I(t) =
Q 0
C
e−t/RC
Q 0
RC
e−t/RC
=
Q^20
RC^2
e−^2 t/RC (383)
PR(t) = VR(t)I(t) =−
Q 0
C
e−t/RC
Q 0
RC
e−t/RC
= −
Q^20
RC^2
e−^2 t/RC (384)
(385)
Note well that the powerdelivered to(+) the circuit by the capacitor equals the powerused by(-)
the resistor!
The final little piece of magic we can look for is energy balance. Suppose we wait a very long
(“infinite”) time – we expect the charge on the capacitor to go to zero in that time. How much
energy appears in the resistor during that entire period?
UR = |/int∞ 0 PR(t)dt|
= Q
(^20)
RC^2
∫∞
0
e−^2 t/RCdt
= −Q
(^20)
2 C
∫∞
0
e−^2 t/RC−^2 dt
RC
= −Q
(^20)
2 C
e−^2 t/RC
∣∣
∣
∞
0
=
Q^20
2 C
(386)
(387)
which justhappensto be the total energy initially on the capacitor: