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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. - PowerPoint PPT Presentation
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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.
Previous assumptions and
how they are changing.
ε S
cR C
ba
R - C in SeriesInitial conditions
An ideal source (r=0)
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 SeriesCharging the Capacitor
εS
cR Cba
i=0
+q
-q
Remember: Current through a
series circuit is the same through all parts.
Total voltage input is equal to the sum of the voltage drops.
At time t=0, Vbc = 0,
So Vab = εTherefore, at t=0,
RR
VI abo
As capacitor C charges,Vbc , so Vab ,Which causes I , so… bcab VV
R - C in SeriesCharging the Capacitor
After a long time C will be fully charged and i = 0, so Vab = 0 and Vbc = ε
During charging…let q be the charge on the capacitor, and i be the current at any time tChoose “+” current for a “+” charge on the left capacitor plate…
At any time t, instantaneous voltages across the resistor and capacitor are…
Vab = iR and Vbc =q/C
εS
cR Cba
i=0
+q
-q
R - C in SeriesUsing Kirchoff’s Law
Then solve for i…
C
qiR
RC
qI
RC
q
RRi C
q
0
εS
cR Cba
i=0
+q
-q
Remember: At t=0, q=0 so initial current I0 = ε/R.
As q , q/RC , capacitor charge approaches final Qf and i to zero.
When i=0, … so… RC
Q
Rf
CQ f
R - C in SeriesNow for the Calculus!
dtRCCq
dq 1
Cq
RCRC
q
Rdt
dqi
1
ε S
cR Cba
i=0 +q
-q
Now integrate both sides…
Rearrange equation to put all terms with q on one side and everything else on the other…
tqdt
RCCq
dq00
1
Use “u-substitution” to evaluate the integral…
RC
t
C
Cq
ln Eliminate the “ln”…
RCt
eC
Cq
R - C in SeriesResults of the Calculus!
RCt
RCt
eIeRdt
dqti o
)(
)1()1()( RCt
RCt
eQeCtq f
ε S
cR Cba
i=0 +q
-q
To find the instantaneous current, remember that i = dq/dt
At any time t during the charging, the instantaneous charge is…
When time t=RC (yes…the product of resistance and capacitance)…
ffe
e
QQq
IIi
632.01
368.01
01
0
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
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.
At t=10τ, I = 0.000045I0 or current is approximately zero.
R-C CircuitsDischarging the capacitor
S
cR C
ba
+Q0
-Q0
Remove battery to discharge capacitor
Assume the capacitor, C, is fully charged, Q0
At t = 0, close the switch…
Initial Conditions at time t=0….
0Qq RC
QI 0
0
and
R-C CircuitsFinding the instantaneous charge during capacitor
discharge
Assume the capacitor, C, is fully charged, Q0
The time-varying current, i, …
RCt
eQtq
RC
t
Q
q
0
0
)(
ln
tq
Qdt
RCq
dqRC
q
dt
dqi
0
10
S
cR C
ba
+q -qii
Now integrate and
solve for q(t)
R-C CircuitsFinding the instantaneous current during capacitor
dischargeThe time-varying charge, q, …
RCt
RCt
RCt
eIeRC
Q
dt
dqti
eQtq
00
0
)(
)(
S
cR C
ba
+q -qii
q
t
Qo
Qo/2Qo/e
RCi
t
Io
Io/2Io/e
RC