So far we considered stationary situations only meaning: dI dt

0, dV dt

0, ...
Next step in generalization is to allow for time dependencies
For consistency we follow the textbook notation where time dependent quantities are labeled by lowercase symbols such as i=I(t), v=V(t), etc
-q +q
V ab
=V a
-V b
V b
V a

What happens during the charging process
The capacitor starts to accumulate charge
A voltage v ab starts to build up which is given by v bc
 q
C
The transportation of charge to the capacitor means a current i is flowing
The current gives rise to a voltage drop across the resistor R which reads: v ab
 iR
Using Kirchhoff’s loop rule we conclude
E  iR
 q

C
0
With the definition of current as i
 dq dt dq
 q

E dt RC R
Inhomogeneous first order linear differential equation
Since in the very first moment when the switch is closed there is not charge in the capacitor we have the initial condition q(t=0)=0

Lets consider common strategies to solve such a differential equation q
We use intuition to get an idea of the solution.
We know that initially q=0 and after a long time when the charging process is finished i=0 and hence
C
E
C
E
Let’s guess t
( )

C
E

1

 e

 t 


( )

C
E

1
 e

 t 

Substitution into dq
 q

E dt RC R
C

E e

 t dq

C dt

E e

 t

C
E
RC


1
 e

 t



C
E
RC
 
RC

Our guess works if we chose  
RC
( )

C
E

1
 e
 t
RC


Guessing is a common strategy to solve differential equations
In the case of a linear first order differential equation there is a systematic integration approach which always works dq
 q

E dt RC R dq

C
E
dt RC q
RC
 q
(

0) dq

C
E
q
  t
0 dt
 with q(t=0)=0  q
0 dq
E

 t
C q RC with C
E
q
  
,
 dq
 ln
C
E
C
E
q

 t
RC


C
E
E dz

 t z RC
C
E
q
C
E
 t
 e RC
( )

C
E

1
 e
 t
RC



q
Q f

C
E
Q f
/ e
 
RC t
After time
(
 
)

C
E
 

1

RC e

1


0.63
C
E 63% of Q f reached
How does the time dependence of the current look like i
( )

C
E

1
 e
 t
RC


I
0

E
R
I
0
/e
 
RC t
 dq

E dt R t e

RC

Now let’s discharge a capacitor
Using Kirchhoff’s loop rule in the absence of an emf we conclude iR q

C
0 dq
 q
 dt RC
0 homogeneous first order linear differential equation
In the very first moment when the switch is closed the current is at maximum and determined by the charge Q(t=0)=Q
0 initially in the capacitor
(

 q
0)
Q
0
 dq q


I
0
 
 
Q
0
1
RC t
RC

0 dt
 ln q
Q
0
  t
RC q i

Q e
 
Q

0
RC t
RC e
 t
RC

 t
RC
Current is opposite to the direction on charging

We close the chapter with an energy consideration for a charging capacitor
When multiplying
E  iR
 q

C
0 i by the time dependent current i
E i
 i R
 iq
C

0
Power delivered by the emf
Rate at which energy is stored in the capacitor
Rate at which energy is dissipate by resistor
Total energy supplied by the battery (emf)
U emf


= E idt
 E

 E
Q
 idt f
 dq
 E
Q f
Total energy stored in capacitor is
U
C
= 

0 q
C idt

Q

0 f q
C dq

1
2
Q
C f
2
2
Half of the energy delivered by the battery is stored in the capacitor no matter what the value of R or C is !
Q f

1
2
U emf

by calculating the energy dissipated in R which must be the remaining half of the energy delivered by the emf
U diss
= 

0
2 i Rdt with 
E
R t e

RC
U diss
=
E 2
R


0
2 t
 e RC dt with 
2 t
RC
 x , dt
 
RC dx
2
U diss
=
-
RC
2
E 2
R


0 x e dx
=
C
E 2
2

0
 x e dx =
C
E 2
2
=
2 C f 
1
2
E
Q f

1
2
U emf

U
C