Download R-C circuits

R-C circuits
So far we considered stationary situations only meaning:
dI
dV
 0,
 0, ...
dt
dt
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
Let’s charge a capacitor again
-q
+q
Vab=Va-Vb
Vb
Va
What happens during the charging process
The capacitor starts to accumulate charge
A voltage vab starts to build up which is given by vbc  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: vab  iR
Using Kirchhoff’s loop rule we conclude
E  iR 
q
0
C
With the definition of current as i 
dq
q
E


dt RC R
dq
dt
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
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 q(t  )  CE
q
CE
t
t
 

Let’s guess q(t )  CE  1  e  


and check if it works and, if so, let’s determine the time constant 
t
 

q(t )  CE  1  e  


dq CE t

e
dt

Substitution into
dq
q
E


dt RC R
CE

e

t

t

CE 


1  e
RC 
 CE

 RC
  RC
Our guess works if we chose   RC
t



q(t )  CE 1  e 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 CE - q

dt
RC
dq
RC 
  dt 
CE - q 0
q ( t 0)
q
with
t
dq
t

0 CE - q RC
q
with q(t=0)=0
CE - q   z, dz  dq
q CE
E
C
dz t

z RC
CE - q t
ln

CE
RC
t
CE - q
 e RC
CE
t



q(t )  CE 1  e RC 


q
Qf / e
  RC
Q f  CE
t
After time   RC
q(t   )  CE 1  e 1   0.63 CE
63% of Qf reached
How does the time dependence of the current look like
t



q(t )  CE 1  e RC 


i
I0 
dq E  RCt
i (t ) 
 e
dt R
E
R
I0/e
  RC
t
Now let’s discharge a capacitor
Using Kirchhoff’s loop rule in the absence of an emf we conclude
iR 
q
0
C
dq
q

0
dt RC
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)=Q0 initially in the capacitor
Q0
RCt
q
dq
1


Q q RC 0 dt
0
I (t  0)  I 0  
ln
q
t

Q0
RC

t
RC
q  Q0e
t

Q0  RCt
i
e
 I 0e RC
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
0
C
i
E i  i2R 
by the time dependent current i
iq
0
C
Rate at which energy
is stored in the capacitor
Power delivered
by the emf
Rate at which energy
is dissipate by resistor
Total energy supplied by the battery (emf)


U emf =  E idt  E  idt  E
0
0
Qf
 dq  E Q
0
Total energy stored in capacitor is

q
U C =  idt 
C
0
Qf

0
2
f
Half of the energy delivered by the battery is stored
in the capacitor no matter what the value of R or C is !
q
1 Qf
1
1
dq 
= E Q f  U emf
C
2 C
2
2
Let’s check this surprising result
by calculating the energy dissipated in R which must be the remaining half of
the energy delivered by the emf

U diss =  i 2 Rdt
with
i (t ) 
dt
with 
0
U diss =
E
2 
R
e

2t
RC
R
e

t
RC
2t
RC
 x , dt  
dx
RC
2
0
2 
RC E
U diss = 2 R
=
E
CE Q f
2C
CE
e
dx
=
0
2
x
2 0
2
C
E
 e dx = 2
x
1
1
 E Q f  U emf  UC
2
2