Transcript Ch. 31 slides

Chapter 31
Fundamentals of Circuits
Phys 133 -- Chapter 31
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Circuits
potentialfield current
Phys 133 -- Chapter 31
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Kirchoff’s rules
KLR: Kirchoff’s loop rule (conservation of energy)
KJR: Kirchoff’s junction rule (conservation of charge)
åI = åI
in
out
Always true!
Phys 133 -- Chapter 31
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Ohm’s Law
--Recall lab: rheostat (resistor), tungsten, carbon
Phys 133 -- Chapter 31
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Question
A typical resistance value is 1Ω. A typical copper wire
has a radius of 0.5 mm.
How long must the typical copper wire be so that it
has a 1Ω resistance?
R=
rL
A
ÞL=
RA
r
copper = 1.7 x 10-8 Ωm
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Ideal wire:
--no potential difference
(“no voltage drop”)
--all potential difference
across resistors
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Circuit elements: real and schematic
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Basic circuit
+
I (+)
≈0
= ( +e) + (-IR) = 0
e
ÞI=
R 133 -- Chapter 31
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≈0
8
Problem A
--Find the magnitude and direction of
the current through the 18 resistor.
On your way to the solution:
•Draw a schematic of this circuit.
•Choose a direction of positive
current
•Indicate the high and low side of
the resistor.
•Apply Kirchoff’s law
--Draw a graph of the potential as a
function of the clockwise distance
around the circuit. Start with V=0 at
lower left corner.
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Problem A (ans)
Draw a schematic
Choose
Indicate
a
the
direction
high and
ofofthis
low
positive
circuit.
side current
of the resistor.
+
-
Apply Kirchoff’s law
I (+)
= ( +e6V ) + (-IR) + (-e 3V ) = 0
1
= A
6
V
6V
3V
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Do Workbook 31.6
a) Iout ? Iin
A. <
B. >
C. =
b) R1 ? R2 ? R3
A.
B.
C.
D.
R1 > R2 > R3
R3 > R1 > R2
R1 = R2 = R3
R2 > R1 > R3
c) E ?
A. Right
B. Left
C. E = 0
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Do Workbook 31.9
a) DV12 = ?
A. 0V
B. 3V
C. <3V
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Resistors-series
I1 = I2 = Iequivalent
I1R1 + I2R2 = Iequivalent Requivalent
Currents are the same
R1 + R2 = Requivalent
In general
Phys 133 -- Chapter 31
Req = R1 + R2 + …
13
Problem B
a) Determine the current through R1 and R2?
b) Determine the potential difference ∆V1 and ∆V2?
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Problem B (ans)
R1 & R2 in series
Requ = R1 + R2
+
I
R1 & R2 in series
1
I1 = I2 = Iequ = A
3
(+e) + (-IR) = 0
ÞI=
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R
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Voltage in a circuit
The light bulb in this circuit has a
resistance of 1 .
What are the values of:
DV12
DV23
DV34
Suppose the bulb is removed
from its socket. Then what are
values of
DV12
DV23
DV34
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A
n
s
w
e
r
s
:
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Resistors-parallel
I1 + I2 = Iequivalent
Potential differences are the same
1
1
1
+
=
R1 R2 Requivalent
In general
Phys 133 -- Chapter 31
1
1 1
= + +…
Req R1 R217
Problem C
a) Determine the current through R1 and R2?
b) Determine the potential difference ∆V1 and ∆V2?
c) What fraction of the current goes through R1?
What happens to this fraction if the value of R2 is
increased?
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Problem C (ans)
R1 & R2 in parallel
1
1
1
= +
Requ R1 R2
+
I
R1 & R2 in parallel
(+e) + (-IR) = 0
ÞI=
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Power and energy
For battery
For resistor
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Problem D
a) How much power is dissipated by each resistor?
b) How much power is provided by each battery?
Series
Parallel
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Problem D (ans)
Series
Parallel
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Real batteries: ideal emf and internal resistance
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Problem E
Compared to an ideal battery, by what percentage does
the battery’s internal resistance reduce the potential
difference across the 30  resistor?
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Problem E (ans)
Compared to an ideal battery, by what percentage does
the battery’s internal resistance reduce the potential
difference across the 30  resistor?
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Conceptual 1
In the circuit below, the battery maintains a constant potential
difference between its terminals at points 1 and 2 (i.e., the
internal resistance of the battery is considered negligible). The
three light bulbs, A, B, and C are identical.
How do the brightnesses of the three
bulbs compare to each other?
A.
B.
C.
D.
A=B=C
B=C>A
A>B=C
A>B>C
*RTP Lab 3 Homework
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Conceptual 2
A wire is connected from the battery terminal at point 1 to
point 4
a) What happens to the potential
difference across bulb B? Explain.
b) What happens to the potential
difference across bulb C? Explain.
c) What happens to the potential
difference between points 1 and 5?
Explain your reasoning.
d) What happens to the brightness of
each of the three bulbs? Explain.
e) What happens to the current through
point 3? Explain.
*RTP Lab 3 Homework
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Resistors
series
parallel
DV
DV1+DV2= DVeq
DV1=DV2
I
I1=I2
I1+I2= Ieq
Req = R1 + R2 + …
1
1
1
= + +…
Req R1 R2
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General relationships
Series
Parallel
DV
add
same
Q, I
same
add
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Problem 31.67
Find the current through and potential difference
across each resistor. (Hint: Collapse and expand.)
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Problem 31.67 (ans)
I3 = I4 = I5 = 1.2A
I1 = I6 = Ibat = 2A
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Problem: Multi-loop circuit
Find the current (I) and potential difference (DV)
across each resistor. (Hint: use KLR and KJR.)
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Circuit problem strategy
--draw currents (any direction), split at each junction
--use KJR at junction(s)
--using KLR, go around a loop (any direction)
Battery--DV increases going (-  +) of battery
--DV decreases going (+  -) of battery
Resistor--DV increases upstream (against current)
--DV decreases downstream (with current)
--solve equations for unknown currents
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Problem F
Find the current (I) and potential difference (DV)
across each resistor. (Hint: use KLR and KJR.)
--draw currents (any direction), split at each junction
--use KJR at junction(s)
--using KLR, go around a loop (any direction)
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Problem F (ans)
+
Apply KLR
-
abcea:
I3
+ I2
I1
-
+
abcdea:
Apply KJR
Junction c:
åI
in
= å Iout
I1 = I2 + I3
ecde:
Junction e:
åI
in
= å Iout
I2 + I3 = I1
5 equations, 3 unknowns
-- Chapter
31
ChoosePhys
3, 133
and
solve
I1 =
4
3
1
A; I2 = A; I3 = A
5
5
5
35
Grounding: convention/safety
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Do Workbook 31.32
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Problem 31.34
Determine the potential at points a through d.
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Problem 31.34 (ans)
Determine the potential at points a through d.
+
-
9V
7V
0V
4V
-
+
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RC circuits: discharge
Before the switch closes
After the switch is closed
Apply KLR
Q(t)
+ (-I(t)R) = 0
C
Q(t)
Þ
= I(t)
RC
dQ(t)
=dt
Þ Q(t) = Q0e
Phys 133 -- Chapter 31
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40
t
RC
RC circuits: discharge
Q(t) = Q0e
-
t
RC
= Q0e
-
t
t
t
dQ(t) Q0 - RCt
I(t) = =
e
= I0e t
dt
RC
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RC circuits: charging
æ
Q(t) = Q0 ç1- e
è
t
RC
- ö
ö
æ
t
=
Q
1e
÷
÷
0ç
ø
è
ø
t
t
dQ(t) Q0 - RCt
I(t) =
=
e
= I0e t
dt
RC
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Problem G
A 20 F capacitor initially charged to 40 C is discharged
through a 5.0 k resistor.
• How long does it take to reduce the capacitor’s charge to
10 C?
• What is the initial current through the resistor?
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Problem G (ans)
Q0 = 40mC
Close switch at t=0, then
Q(t) = Q0e
-
t
RC
= Q0e
-
= -I0 R
t
t
Þ I0 =
Q0
RC
Find t when Q(t)=20µC
-
t 20
1s
20mC = (40mC)e
t
- 20
1
æ 1ö
1s
Þ t20 = - ln
= 0.693s
Þ =e
è
ø
2
2
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