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PHYS 222
Worksheet 13 – RC Circuits
Supplemental Instruction
Iowa State University
Alek Jerauld
PHYS 222
Dr. Paula Herrera-Siklódy
2/20/12
Useful Equations
I  I0et / RC 
 et / RC
R
I  I0et / RC  
  RC
Q0 t / RC
e
RC
Current through capacitor during capacitor charging
Current through capacitor during capacitor discharging
Time constant. This is the time it takes for the current to
reach 36.8% of its original current
Rules
 Capacitor acts like a straight wire when it is connected to the circuit.

Capacitor acts like a break after
∞ amount of time (a very long time).
Diagrams
The above equations only apply to this type of circuit:
In order to find current through other types of circuits, you must convert the other circuit into
this form.
Related Problems
1) A 4.40 µF capacitor that is initially uncharged is connected in series with a 6.60 kΩ
resistor and an emf source with ε = 130 V negligible internal resistance. (Book 26.38)
(a) Just after the circuit is completed, what is the voltage drop across the capacitor?
Capacitor acts like a straight wire, thus V = 0 V
(b) Just after the circuit is completed, what is the voltage drop across the resistor?
Since no voltage drop occurs across the capacitor, all the voltage drops across the resistor:
Vr = 130 V
(c) Just after the circuit is completed, what is the current through the resistor?
V
I  R  0.0197 A
R
(d) A long time after the circuit is completed (after many time constants) what is the voltage
drop across the capacitor?
After a long time, capacitor acts like a very large resistor (or break). So voltage across the
capacitor becomes Vc = 130 V
(e) A long time after the circuit is completed (after many time constants) what is the voltage
drop across the resistor?
Since all the voltage drops across the capacitor after a very long time, Vr = 0 V
(f) A long time after the circuit is completed (after many time constants) what is the current
through the resistor?
Again, capacitor acts line a break or a giant resistor, using ohm’s law with R = infinity, I = 0
A.
2) In the circuit shown in the figure each capacitor
initially has a charge of magnitude 3.60 nC on its
plates. After the switch S is closed, what will be the
current in the circuit at the instant that the capacitors
have lost 80.0% of their initial stored energy? (Book
26.45)
1
Ceq 
 4.62 pF
1
1
1


C1 C2 C3
I  I0et / RC 
Q0 t / RC
Q
e
  0 e ln(5)/2  13.9 A
RCeq
RCeq
3) In the circuit in the figure the capacitors are all
initially uncharged, the battery has no internal
resistance, and the ammeter is idealized. (Book
26.47)
(a) Find the reading of the ammeter just after the
switch S is closed.
When the switch is initially closed all capacitors
act like straight wires:
Then you can eliminate components that are in parallel with a straight wire. The circuit
looks like this:
Req  75  15 
I
1
1
1

50 25
 106.67 
V
 0.937 A
Req
(b) Find the reading of the ammeter after the switch has been closed for a very long time.
After a long time, the capacitors act like breaks in the circuit:
Req  75  15  25  25  25  165 
I
V
 0.606 A
Req
4) In the circuit shown in the figure both capacitors are initially charged to 50.0 V. (Book
26.41)
(a) How long after closing the switch S will the potential across each capacitor be reduced
to 15.0 V?
Ceq  35 F
Req  80 
i  Ioe
Io 

 t / Req Ceq
V0
v
,i
Req
Req
 v 
v V0 t / ReqCeq
 e
 t   ln   Req Ceq  3.37 ms
R R
 V0 
(b) What will be the current at that time?
v
i   0.188 A
R