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Today’s agenda:
Measuring Instruments: ammeter, voltmeter, ohmmeter.
You must be able to calculate currents and voltages in circuits that contain “real”
measuring instruments.
RC Circuits.
You must be able to calculate currents and voltages in circuits containing both a resistor
and a capacitor. You must be able to calculate the time constant of an RC circuit, or use
the time constant in other calculations.
RC Circuits
RC circuits contain both a resistor R and a capacitor C (duh).
Until now we have assumed that charge is
instantly placed on a capacitor by an emf.
Q
t
The approximation resulting from this
assumption is reasonable, provided the
resistance between the emf and the capacitor
being charged/discharged is small.
If the resistance between the emf and the
capacitor is finite, then the capacitor does not
change instantaneously.
Q
t
Charging a Capacitor
Switch open, no current flows.
I
+q +
-q
Close switch, current flows.
q
- IR = 0
C
C
 +-
Apply Kirchoff’s loop rule*
(green loop) at the instant
charge on C is q.
ε-
-
This equation is
deceptively
complex because
I depends on q
and both depend
on time.
R
switch
t<0
t>0
*Convention for capacitors is “like” batteries: negative if going across from + to -.
Limiting Cases
q
ε - - IR = 0
C
When t=0, *q=0 and I0=/R.
When t is “large,” the capacitor
is fully charged, the current
“shuts off,” and Q=C.
I
+
-
=0
C qF0=C
 +R IF=0
t=
t=0
switch
*q=0 at t=0 only if the capacitor is initially uncharged (read the problems carefully!)
 = IR is true only at time t=0 or when q=0! VR = IR is always true, but VR is the potential
difference across the resistor, which you may not know. Using V = IR to find the voltage
across the capacitor is likely to lead to mistakes unless you are very careful.
Math:
ε-
q
- IR = 0
C
ε q
I= R RC
dq ε q
Cε q
Cε - q
= =
=
dt R RC RC RC
RC
dq
dt
=
Cε - q RC
dq
dt
=q- Cε
RC
More math:

q
0
t dt
dq
=-
0 RC
q- Cε
1 t
ln  q - Cε  0 = dt

0
RC
q
t
 q- Cε 
ln 
=

-C
ε
RC


t
q- Cε
= e RC
-Cε
q- Cε = -Cε e
-
t
RC
Still more math:
q= Cε - Cε e
-
t
RC
t


RC
q = Cε 1- e 


t


RC
q  t  = Q 1- e 


Why not just
solve this for
q and I?
ε-
q
- IR = 0
C
 1 - RCt  Cε - RCt
dq
ε - RCt
ε - t
I  t  = = Cε 
e =
e = e = e
dt
R
R
 RC
 RC
 = RC is the “time constant” of the circuit; it tells us “how
fast” the capacitor charges and discharges.
recall that this is I0,
also called Imax
Charging a capacitor; summary:
t


RC
q  t  = Qfinal 1- e 


ε - RCt
It = e
R
Charging Capacitor
0.01
0.05
0.008
0.04
0.006
0.03
I (A)
q (C)
Charging Capacitor
0.004
0.002
0.02
0.01
0
0
0
0.2
0.4
0.6
t (s)
0.8
1
0
0.2
0.4
0.6
0.8
t (s)
Sample plots with =10 V, R=200 , and C=1000 F.
RC=0.2 s
1
In a time t=RC, the capacitor charges to Q(1-e-1) or 63% of its
capacity…
…and the current drops to Imax(e-1) or 37% of its maximum.
Charging Capacitor
0.01
0.05
0.008
0.04
0.006
0.03
I (A)
q (C)
Charging Capacitor
0.004
0.002
0.02
0.01
0
0
0
0.2
0.4
0.6
0.8
1
t (s)
0
0.2
0.4
0.6
0.8
t (s)
RC=0.2 s
=RC is called the time constant of the RC circuit
1
Discharging a Capacitor
Capacitor charged, switch
open, no current flows.
Close switch, current flows.
Apply Kirchoff’s loop rule*
(green loop) at the instant
charge on C is q.
q
- IR = 0
C
I
C
+Q
+q
-q
-Q
R
switch
t<0
t>0
*Convention for capacitors is “like” batteries: positive if going across from - to +.
Math:
q
- IR = 0
C
IR =
q
C
I=
dq
dt
-R
dq q
=
dt C
dq
dt
=q
RC
negative because
charge decreases
More math:
t dt
dq
1 t
Q q = - 0 RC = - RC 0 dt
q
1 t
ln  q  Q = dt

0
RC
q
t
q
ln   = RC
Q
q(t) = Q e
-
t
RC
t
dq Q - RCt
I(t) = - =
e = I0 e RC
dt RC
same equation
as for charging
Disharging a capacitor; summary:
q(t) = Q0 e
-
t
RC
I  t  = I0 e
0.01
0.05
0.008
0.04
0.006
0.03
0.004
0.02
0.002
0.01
0
0
0
0.2
0.4
0.6
t (s)
t
RC
Discharging Capacitor
I (A)
q (C)
Discharging Capacitor
-
0.8
1
0
0.2
0.4
0.6
0.8
t (s)
Sample plots with =10 V, R=200 , and C=1000 F.
RC=0.2 s
1
In a time t=RC, the capacitor discharges to Qe-1 or 37% of its
capacity…
…and the current drops to Imax(e-1) or 37% of its maximum.
Discharging Capacitor
0.01
0.05
0.008
0.04
0.006
0.03
I (A)
q (C)
Discharging Capacitor
0.004
0.02
0.002
0.01
0
0
0
0.2
0.4
0.6
0.8
1
t (s)
0
0.2
0.4
0.6
t (s)
RC=0.2 s
0.8
1
Notes
ε - RCt
It = e
R
I  t  = I0 e
-
t
RC
This is for charging a capacitor.
/R = I0 = Imax is the initial current,
and depends on the charging emf
and the resistor.
This is for discharging a capacitor.
I0 = Q/RC, and depends on how
much charge Q the capacitor
started with.
I0 for charging is equal to I0 for discharging only if the
discharging capacitor was fully charged.
Notes
In a series RC circuit, the same current I flows through both
the capacitor and the resistor. Sometimes this fact comes in
handy.
In a series RC circuit, where a source of emf is present (so
this is for capacitor charging problems)...
VR + VC = V
VR = V - VC = IR
V - VC
I =
R
Vc and I must be at the same instant in time for this to work.
Any technique that begins with a starting equation and is worked
correctly is acceptable, but I don’t recommend trying to memorize a
bunch of special cases. Starting with I(t) = dq(t)/dt always works.
Notes
In a discharging capacitor problem...
VR = VC = IR
VC
I=
R
So sometimes you can “get away” with using V = IR, where V
is the potential difference across the capacitor (if the circuit
has only a resistor and a capacitor).
Rather than hoping you get lucky and “get away” with using
V = IR, I recommend you understand the physics of the
circuit!
Homework Hints
Q(t) = CV(t)
This is always true for a capacitor.
t


RC
q  t  = Qfinal 1- e 


Qfinal = C, where  is the potential
difference of the charging emf.
q(t) = Q0 e
V = IR
-
t
RC
Q0 is the charge on the capacitor
at the start of discharge. Q0 = C
only if you let the capacitor charge
for a “long time.”
Ohm’s law applies to resistors, not
capacitors. Can give you the
current only if you know V across
the resistor. Safer to take dq/dt.