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Lecture 10, ACT 1
• Consider the circuit shown:
1A
50W
a
– What is the relation between Va -Vd
and Va -Vc ?
b
12V
(a) (Va -Vd) < (Va -Vc)
(b) (Va -Vd) = (Va -Vc)
(c) (Va -Vd) > (Va -Vc)
1B
1B
(b) I1 = I2
20W
80W
d
– What is the relation between I1 and I2?
(a) I1 < I2
I2
I1
(c) I1 > I2
c
Lecture 10, ACT 1
• Consider the circuit shown:
1A
50W
a
– What is the relation between Va -Vd
and Va -Vc ?
(a) (Va -Vd) < (Va -Vc)
(b) (Va -Vd) = (Va -Vc)
(c) (Va -Vd) > (Va -Vc)
b
I2
I1
12V
20W
80W
d
c
• Do you remember that thing about potential being independent of path?
Well, that’s what’s going on here !!!
(Va -Vd) = (Va -Vc)
Point d and c are the same, electrically
Lecture 10, ACT 1
• Consider the circuit shown:
1A
50W
a
– What is the relation between Va -Vd
and Va -Vc ?
b
12V
(a) (Va -Vd) < (Va -Vc)
(b) (Va -Vd) = (Va -Vc)
(c) (Va -Vd) > (Va -Vc)
1B
1B
20W
• Note that: Vb -Vd
• Therefore,
(b) I1 = I2
(c) I1 > I2
= Vb -Vc
I1 (20W)  I 2 (80W)
80W
d
– What is the relation between I1 and I2?
(a) I1 < I2
I2
I1
I1  4I 2
c
Summary of Simple Circuits
• Resistors
in series: Reffective  R1  R2  R3  ...
Current thru is same;
• Resistors
1
in parallel: R
effective
Voltage drop across is IRi

1 1 1
   ...
R1 R2 R3
Voltage drop across is same;
Current thru is V/Ri
Kirchhoff’s laws: (Tipler problems on Kirchhoff’’s rules)
V
n
0
loop
I in   I out
Batteries
(“Nonideal” = cannot output arbitrary current)
I
I
• Parameterized with
R
"internal resistance"
V  e  Ir
r
V
e
e  Ir  IR  0
e
I
Rr

R
V e
Rr
Internal Resistance Demo
As # bulbs increases, what happens to “R”??
I
r
V
R
e
How big is “r”?
Power
Batteries & Resistors
Energy expended
chemical
to electrical
to heat
Rate is:
energy
 power  sJ 
time
What’s happening?
Assert:
Charges per time
P  VI
Potential difference per charge
or you can write it as
For Resistors:
Units okay?
P  IR I  I 2 R
P  V V R   V 2 R
Joule  Coulomb J  Watt
Coulomb second s
More complex now…
I
Let’s try to add a Capacitor to
our simple circuit
I
R
Recall voltage “drop” on C?
V
C
e
Q
C
Q
Write KVL:
e  IR 
0
C
What’s wrong here?
Consider that I 
dQ
dt
and substitute. Now eqn has only “Q”.
KVL gives Differential Equation !
e R
dQ Q

0
dt
C
The Big Idea
• Previously:
– Analysis of multi-loop circuits with batteries and
resistors.
– Main Feature: Currents are attained instantaneously
and do not vary with time!!
• Now:
– Just added a capacitor to the circuit.
– What changes??
• KVL yields a differential equation with a term
proportional to Q and a term proportional to
I = dQ/dt.
dQ Q
eR
 0
dt C
The Big Idea
dQ Q
eR
 0
dt C
• Physically, what’s happening is that the final charge
cannot be placed on a capacitor instantly.
• Initially, the voltage drop across an uncharged
capacitor = 0 because the charge on it is zero !
(V=Q/C)
• As current starts to flow, charge builds up on the
capacitor, the voltage drop is proportional to this
charge and increases; it then becomes more
difficult to add more charge so the current slows
2
Lecture 10, ACT 2
2A
• At t=0 the switch is thrown
from position b to position a in
the circuit shown: The
capacitor is initially
uncharged.
a
2B
(b) I0+ = e /2R
b
(c) I0+ = 2e /R
long time?
(b) I = e /2R
C
e
– What is the value of the current I after a very
(a) I = 0
I
R
– What is the value of the current I0+
just after the switch is thrown?
(a) I0+ = 0
I
(c) I > 2e /R
R
Lecture 10, ACT 2
2A
• At t=0 the switch is thrown
from position b to position a in
the circuit shown: The
capacitor is initially
uncharged.
a
(b) I0+ = e /2R
I
R
b
C
e
– What is the value of the current I0+
just after the switch is thrown?
(a) I0+ = 0
I
R
(c) I0+ = 2e /R
•
Just after the switch is thrown, the capacitor still has no
charge, therefore the voltage drop across the capacitor = 0!
•
Applying KVL to the loop at t=0+, e IR  0  IR = 0  I = e /2R
Lecture 10, ACT 2 a
2A
• At t=0 the switch is thrown from
position b to position a in the circuit
shown: The capacitor is initially
uncharged.
– What is the value of the current I0+
just after the switch is thrown?
I
I
R
b
C
e
R
(a) I0+ = 0
2B
(b) I0+ = e /2R
(c) I0+ = 2e /R
– What is the value of the current I after a very
long time?
(a) I = 0
(b) I = e /2R
(c) I > 2e /R
• The key here is to realize that as the current continues to flow, the
charge on the capacitor continues to grow.
• As the charge on the capacitor continues to grow, the voltage across
the capacitor will increase.
• The voltage across the capacitor is limited to e ; the current goes to 0.
Behavior of Capacitors
• Charging
– Initially, the capacitor behaves like a wire.
– After a long time, the capacitor behaves like an open switch.
• Discharging
– Initially, the capacitor behaves like a battery.
– After a long time, the capacitor behaves like a wire.
Discharging Capacitor
• The capacitor is initially fully
charged, Q = Q0. At t = 0 the switch
is thrown from position a to
position b in the circuit shown.
• From KVL:
R
dQ
dt

dQ

dt
Q
0
C
1
RC
I
a
b
Q(t )  e
Therefore,
e
Q
dQ
 ae at  aQ
dt
where
From initial condition, Q(0) = Q0, we get:
Q(t )  Q0e
t
RC
R
+ +
C
Q(t) must be an exponential function
of the form:

at
I
1
a
RC
- -
Summary
• Kirchhoff’s Laws
– KCL: Junction Rule (Charge is conserved)
– Review KVL (V is independent of path)
• Non-ideal Batteries & Power
• Discharging of capacitor through a Resistor:
Q (t )  Q0 e
t
RC
Reading Assignment: Chapter 26.6
Examples: 26.17,18 and 19
Two identical light bulbs are
represented by the resistors
R2 and R3 (R2 = R3 ). The
switch S is initially open.
2) If switch S is closed, what happens to the brightness of the bulb R2?
a) It increases
b) It decreases
c) It doesn’t change
3) What happens to the current I, after the switch is closed ?
a) Iafter = 1/2 Ibefore
b) Iafter = Ibefore
c) Iafter = 2 Ibefore
I
Four identical resistors are
connected to a battery as
shown in the figure.
R2
R1
R4
E
R3
5) How does the current through the battery change after the
switch is closed ?
a) Iafter > Ibefore
b) Iafter = Ibefore
c) Iafter < Ibefore
Before:
Rtot = 3R
Ibefore = 1/3 E/R
After:
R23 = 2R
R423 = 2/3 R
Rtot = 5/3 R
Iafter = 3/5 E/R
Appendix: A three-loop KVL example
• Identify all circuit nodes these are where KCL eqn’s
are found
I3
I1
I2
– determine which KCL
equations are algebraically
independent (not all are in
this circuit!)
–
–
–
–
I1=I2+I3
I4=I2+I3
I4=I5+I6
I1=I5+I6
I1=I4
I1=I2+I3
I4+I5+I6
• Analyze circuit and identify
all independent loops where
S DVi = 0 < KVL
I6
I4
I5
A three-loop KVL example
• Here are the node equations from
applying Kirchoff’s current law:
I1=I4
I1=I2+I3
I4+I5+I6
• Now, for Kirchoff’s voltage law:
(first, name the resistors)
I6R6+I5R5=0
I2R2b+I2R2a- I3R3 =0
VB-I1R1-I2R2a- I2R2b-I4R4-I5R5 = 0
Six equations, six unknowns….
I3
R1
I1
I2
R3
R2b
I6
R6
R5
• There are simpler ways of
analyzing this circuit, but this
illustrates Kirchoff’s laws
R2a
I5
R4
I4