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Announcements
• Quiz II March 3rd
– Median 86; mean 85
• Quiz III: March 31st
• Office Hrs: Today
– 2-3pm
A Multiloop Circuit
I1 + I3 = I2
1.5 – 3I2 = 0
9 – 5I1 – 3I2 = 0
I2 = 1.5/3 = 0.5 A
I1 = (9 – 3I2)/5 = 1.5 A
I3 = I2 – I1 = 0.5 – 1.5 = – 1 A
I1
–
+
9V
5
I2
3
1.5 V
– +
I3
What
Whatisisthe
thevoltage
conservation
loop rule
of
you
current
get applied
law associated
to the upper
with
loop?
the junction on the right?
A)
A) 9I1++5II21 =+ I3I
32=0
B)
B) 9I+1 +5II13 –=3I
I22 = 0
C)
C) 9I–2 +5II13+=3I
I12 = 0
D)
D) 9I1–+5II21 +
– 3I
I3 2==00
A Multiloop Circuit
•There is one more loop in the
problem.
9 – 5I1 -1.5 = 0
I1 = (9 – 1.5)/5 = 1.5 A
I1
–
+
9V
5
I2
3
1.5 V
– +
I3
•We only had one resistor and so only had to consider
one current. This can simplify problems!
Odd Circuit
What is the current through the
resistor?
A) 3.6 A
B) 1.8A
C) 90 A
D) 0 A
–
+
9V
5
9V
– +
Each of the resistors in the diagram is 12 .
The resistance of the entire circuit is:
A)120
B) 25
C) 48
D) 5.76
RC circuits: Prior to Steady-State
•Thus far we have been referring to circuits in
which the current does not vary in time, i.e.,
steady-state circuits
•When we mix capacitors and resistors, the
currents can vary with time?
Why?!
We need to charge the capacitor!
•A capacitor which is being charged conducts
like a wire
•After charging, the capacitor acts like a broken
wire
E
–
+
S1
C
R
S2
RC circuits: Prior to Steady-State
•Recall: the voltage across a capacitor is: V=q/C
E
–
+
S1
•When the capacitor is fully charged the voltage
is e ( e.g. it acts like a broken wire)
•Prior, the voltage is V, i.e. there is a voltage
drop.
Close S1
Apply the loop rule:
C
R
q
e  iR   0
c
dq
q
e
R 0
dt
c
The result is a differential
equation.
RC circuits: differential Eqns
dq
q
e
R 0
dt
c
q  qb  Ke
 t
Plausibility argument:
dq
q
R 0
dt
c
Differential equation.
General Solution:
qb and K are determined
from boundary conditions
and  from the parameters
of the differential
equation
dq  dt

q
Rc
RC circuits: Differential Eqns
dq
q
R 0
dt
c
dq  dt

q
Rc
q  Ke
t
Rc
Plausibility argument:
Integrate both sides to solve:
K is determined from
boundary conditions
More general equation and solution:
dq
q
e
R 0
dt
c
q  qb  Ke
 t
RC circuits: Boundary Conditions
q  qb  Ke
 t
Charging:
0  qb  K
At t=0, q=0
As t goes to infinity, q=eC
eC  qb
Combining these together
and:
q  Ce  Cee
As an exercise do the same for discharging
t
RC
RC circuits
•Capacitor/resistor systems charge or discharge over time
Charging:



q(t )  Qmax 1  e t / RC  Qmax 1  e t /

 is the time constant, and equals RC.
Discharging:
q(t )  Qe
t /
Qualitatively: RC controls how long it takes to charge/discharge
completely. This depends on how much current can flow (R)
and how much charge needs to be stored (C)
As an exercise, show that RC has units sec
RC circuits: Discharging
•Circuit with battery, resistor, and capacitor
•Switch S1 is closed, then opened
•At t = 0, switch S2 is closed
•What happens?
•Battery increases voltage on capacitor to V = E
•Charge Q = CV is stored on capacitor
•At t=0. Current begins to flow
dQ
I 
dt
Q(t )  Q0e
 t /( RC )
What is the current? [exercise for the class]
E
–
+
S1
C
– +
R
S2
Time Constants
•Time constants are common in science!
Given a time constant, t, how long does one have to wait for
something to decay by:
10%
.105
25%
.288
50%
.693
90%
2.30
99%
4.60
99.99%
9.21
Four circuits have the form shown in the diagram. The capacitor is
initially uncharged and the switch S is open.
The values of the emf , resistance R, and the capacitance C for each of
the circuits are
circuit 1:
18 V, R = 3 , C = 1 µF
circuit 2:
circuit 3:
18 V, R = 6 , C = 9 µF
12 V, R = 1 , C = 7 µF
circuit 4:
10 V, R = 5 , C = 7 µF
Which circuit has the largest current right after the switch is closed?
Which circuit takes the longest time to charge the capacitor to
½ its final charge?
Which circuit takes the least amount of time to charge the capacitor to
½ its final charge?
In the figure below, resistor R3 is a variable resistor and the battery
is an ideal 18 V battery. Figure 28N-2b gives the current I through
the battery as a function of R3. (The vertical axis is marked in
increments of 2.5 mA and the horizontal axis is marked in
increments of 3.0 .) The curve has an asymptote of 5.0 mA as
R3 goes to infinity. (Think of the wire and bulb quiz )
When R3-> infinity, we can ignore R3
V  I ( R1  R 2)  18V  .005 A( R1  R 2)
( R1  R 2)  3600
In Fig. 28N-2a, resistor R3 is a variable resistor and the battery is
an ideal 18 V battery. Figure 28N-2b gives the current i through
the battery as a function of R3. (The vertical axis is marked in
increments of 2.5 mA and the horizontal axis is marked in
increments of 3.0 .) The curve has an asymptote of 5.0 mA as
R3 goes to infinity. (Think of the wire and bulb quiz )
When R3-> 0, we can use the loop rule without R2
V  I ( R1)  18V  .015 A( R1)
(R1)  1200
(R 2)  2400