Transcript RC Circuits - Humble ISD

Resistors and Capacitors in the Same Circuit!!
R-C CIRCUITS
Previous assumptions and
how they are changing.
So far we have assumed resistance (R),
electromotive force or source voltage (ε),
potential (V), current (I), and power (P) are
constant.
When charging or discharging a capacitor I, V,
P change with time, we will use lower-case i,
v, and p for the instantaneous values of
current, voltage, and power.
An ideal source
(r=0)
R - C in Series
Initial conditions
ε
a
S
b
R
c
C
At time t=0 we will close the switch. The instantaneous
current at this time is i=0 and the instantaneous charge
on the capacitor is q = 0 because at this instant there has
been no chance for charge to build on the capacitor…yet!
R - C in Series
Charging the Capacitor
ε
S
i=0
a
R
+q -q
b
C
At time t=0, Vbc = 0,
So Vab = ε
Therefore, at t=0,
Vab 
Io 

R
R
c
Remember:
Current through a series
circuit is the same
through all parts.
Total voltage input is
equal to the sum of the
voltage drops.
As capacitor C charges,
Vbc , so Vab ,
Which causes I , so…
Vab  Vbc  
R - C in Series
Charging the Capacitor
ε
S
i=0
a
R
After a long time C will be
fully charged and i = 0, so
Vab = 0 and Vbc = ε
+q -q
b
C
c
During charging…
let q be the charge on the capacitor,
and i be the current at any time t
At any time t, instantaneous
voltages across the resistor
and capacitor are…
Choose “+” current for a “+”
charge on the left capacitor plate…
Vab = iR and Vbc =q/C
R - C in Series
Using Kirchoff’s Law
ε
S
i=0
a
+q -q
b
R
q
  iR 
C
i
Then solve for i…
c
C
  qC
R

q
q
 
 I0 
R RC
RC
Remember: At t=0, q=0 so initial current I0 = ε/R.
As q
, q/RC
When i=0, …

R
, capacitor charge approaches final Qf and i

Qf
RC
so…
Q f  C
to zero.
ε
S
i=0
a
R
R - C in Series
Now for the Calculus!
+q -q
b
C
c
dq 
q
1
q  C 
i
 

dt
R RC RC
dq
1
 dt
q  C RC
q
dq
1 t
0 q  C  RC 0 dt
Rearrange equation to put all
terms with q on one side and
everything else on the other…
Now integrate both sides…
Use “u-substitution”
to evaluate the
integral…
 q  C
ln 
  C
 t

 RC
Eliminate
the “ln”…
t
q  C
 e RC
 C
R - C in Series
Results of the Calculus!
ε
S
i=0
a
+q -q
b
R
C
c
At any time t during the charging, the instantaneous charge is…
q(t )  C (1  e
t
RC
)  Q f (1  e
t
RC
)
To find the instantaneous current, remember that i = dq/dt
t
dq  t RC
i(t )   e  I o e RC
dt R
R - C in Series
The Graphs
i
q(t )  Q f (1  e
Io
Io/2
Io/e
q
t
RC
i(t )  I oe
t
Qf
Qf/e
Qf/2
RC
RC
t
RC
)
What is so special about time = RC?
When time t=RC (yes…the product of resistance and capacitance)…
i  I 0 1e   0.368I 0
q  1  1 e Q f  0.632Q f
Or about 37% of the original current
Or about 63% of the original voltage
The product of R and C is called the time constant of the
circuit. The time constant is a measure of how fast the
capacitor charges. The symbol for time constant is τ (greek
letter “tau”). It is calculated as RC because a that time the
exponent of the “e” function becomes “-1” which gives us
the equations above.
  RC
At t=10τ, I = 0.000045I0 or current is approximately zero.
NOTE:
If the time constant is small, the capacitor will charge quickly.
If the resistance is small, current is small, so the capacitor charges more quickly.
If the time constant is large, it will take more time for the capacitor to charge.
Remove battery to
discharge capacitor
R-C Circuits
Discharging the capacitor
S
Assume the capacitor,
C, is fully charged, Q0
+Q0 -Q0
a
b
R
C
c
Initial Conditions at time t=0….
At t = 0, close the switch…
q  Q0
and
 Q0
I0 
RC
R-C Circuits
Finding the instantaneous charge during capacitor discharge
S
i
i
a
R
+q
b
Assume the capacitor,
C, is fully charged, Q0
-q
C
c
The time-varying current, i, …
dq  q
i

dt RC
q dq
1 t
Q0 q  RC 0 dt
Now
integrate
and solve
for q(t)
 q  t
ln   
 Q0  RC
q(t )  Q0 e
t
RC
R-C Circuits
Finding the instantaneous current during capacitor discharge
S
The time-varying charge, q, …
i
i
a
R
RC
+q
b
q (t )  Q0 e
-q
C
t
RC
t
dq  Q0 t RC
i (t ) 

e  I 0 e RC
dt
RC
c
t
q
Qo
Io/e
Io/2
Qo/2
Qo/e
Io
i
RC
t
R-C Circuits
Energy & Power
During charging…
The power of the battery, resistor and capacitor are as follows…
Pbattery  i
Presistor  i 2 R
Pcapacitor  iVbc  i q C
So…the total power of the circuit is…
i  i R  i
2
q
C
Where ½ the energy is stored in the capacitor and the other ½ is dissipated by the resistor.