Course Updates
http://www.phys.hawaii.edu/~varner/PHYS272-Spr10/physics272.html
Notes for today:
1) Complete Chap 26, Problem session Friday
2) Assignment 6 (Mastering Physics) online and
separate, written problems due Monday
3) Quiz 3 on Friday
4) Review Midterm 1
Kirchhoff’s Rules
Kirchhoff’s rules are statements used to solve for currents and
voltages in complicated circuits. The rules are
Rule I. Sum of currents into any junction
is zero.
∑ Ii = 0
i
I1 + I 2 = I12
Why? Since charge is conserved.
Rule II. Sum of potential differences
in any loop is zero. (This includes emfs)
a
∑Vi = 0
i
Why? Since potential (energy)
is conserved
d
b
c
Vab + Vbc + Vcd + Vda = 0
RC Circuits
I
a
I
a
I
I
R
C
ε
Cε
R
b
b
RC
+ +
C
ε
2RC
Cε
RC
- -
2RC
q = Cεe − t / RC
(
q
0
q = Cε 1 − e − t / RC
t
)
q
0
t
RC Circuits
(Time-varying currents, charging)
•
I
a
Charge capacitor:
I
R
C initially uncharged;
connect switch to a at t=0
Calculate current and
b
C
ε
charge as function of time.
Q
ε − IR − = 0
C
•Convert to differential equation for Q:
• Loop theorem ⇒
dQ
I=
dt
⇒
Would it matter where R
is placed in the loop??
dQ Q
ε =R +
dt C
Charging Capacitor
•
I
a
Charge capacitor:
I
R
ε=R
dQ Q
+
dt C
b
C
ε
• Guess solution:
Q = Cε (1 − e
−t
RC
)
•Check that it is a solution:
dQ
= C ε e − t / RC
dt
⇒
Note that this “guess”
fits the boundary
conditions:
1 ⎞
⎛
−
⎜
⎟
⎝ RC ⎠
−t
dQ Q
−t / RC
R
+ = −εe
+ ε (1 − e RC ) = ε
dt C
!
t = 0 ⇒Q = 0
t = ∞ ⇒ Q = Cε
Charging Capacitor
•
Q = Cε (1− e
−t / RC
•
I
R
)
b
C
ε
Current is found from
differentiation:
dQ ε −t / RC
I=
= e
dt R
I
a
Charge capacitor:
⇒
Conclusion:
• Capacitor reaches its final
charge(Q=Cε ) exponentially
with time constant τ = RC.
• Current decays from max
(=ε /R) with same time
constant.
Charging Capacitor
Charge on C
Q = Cε (1 − e− t / RC )
Max = Cε
RC
2RC
Cε
Q
63% Max at t = RC
0
Current
I =
dQ ε − t / RC
= e
dt
R
Max = ε /R
t
ε /R
I
37% Max at t = RC
0
t
Discharging Capacitor
I
a
dQ Q
R
+ =0
dt C
b
I
R
C
ε
• Guess solution:
+ +
Q = Q 0 e − t /τ = C ε e − t / RC
• Check that it is a solution:
dQ
= C ε e − t / RC
dt
⇒
Note that this “guess”
fits the boundary
conditions:
1 ⎞
⎛
−
⎜
⎟
⎝ RC ⎠
dQ Q
−
−
R dt + = − ε e t / RC + ε e t / RC = 0
C
!
t = 0 ⇒ Q = Cε
t =∞⇒Q=0
- -
Discharging Capacitor
• Discharge capacitor:
Q = Q 0 e − t /τ = C ε e − t / RC
• Current is found from
differentiation:
dQ
ε −t / RC
I=
=− e
dt
R
⇒
Minus sign:
Current is opposite to
original definition,
i.e., charges flow
away from capacitor.
I
a
b
I
R
+ +
C
ε
- -
Conclusion:
• Capacitor discharges
exponentially with time constant
τ = RC
• Current decays from initial max
value (= -ε/R) with same time
constant
Discharging Capacitor
Cε
Charge on C
Q = C ε e − t / RC
Max = Cε
RC
2RC
Q
37% Max at t = RC
0
zero
t
0
Current
dQ
ε
I =
= − e − t / RC
dt
R
I
“Max” = -ε/R
37% Max at t = RC
-ε /R
t
Midterm 1 statistics
B
F
D
C
A
Cumulative statistics
Midterm Review
• Not many systematic, consistent problems
• However, some recurrent themes:
–
–
–
–
Electric Potential (V) vs. Potential Energy (U)
Electric field sums as vector quantity
Potential field (V) sums as a scaler
Area of a circle is πr2
• Detailed points breakdown (by roster ID) linked
from course webpage
• Will go over in outline form problems next…
For next time
• HW #6 Assigned Æ due next Monday
• Quiz #3 on Friday
• Problem session Friday prior to quiz