Exam 1:
Exam 1:
*Covers all readings, lectures, homework from Chapters
17 through 20.4
*Exam 1: Thursday September 29, 2009 from 8 PM - 10
PM
*The exam will be multiple choice and is meant to be done
within 75 minutes by a well-prepared student. We will
give 120 minutes starting promptly at the listed time, so
please be on time!
* Room PHYS 112 for last names A through K
* Room PHYS 114 for last names L through Z
Be sure to bring your student ID card,
calculator, pencil and your own one-page
(two-side) crib sheet.
NOTE THAT FEW EQUATIONS WILL BE GIVEN – YOU ARE REMINDED
THAT IT IS YOUR RESPONSIBILITY TO CREATE WHATEVER TWO-SIDED
CRIB SHEET YOU WANT TO BRING TO THIS EXAM.
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1
Circuit Analysis
2!
4!
8!
1.5 V
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Circuit Analysis
How much current flows through the
2 ! resistor?
8!
2. Find current:
10 !
1. Simplify
a) replace parallel resistors
1.5 V
2!
4!
0.15A2 !
b) replace series resistors
4!
4!
8!
1.5 V
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2
8!
1.5 V
3
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4
Power in Electrical Circuits
Power in Electrical Circuits
Current I through resistor:
Power in electrical
circuit:
charges move through potential difference "V
P = I!V
Potential energy change: !PEE = q!V
Units: A ! V =
I
"V
Power = rate of energy conversion, or work per unit
time.
Power dissipated in any circuit is the product of current flowing
through that circuit and potential difference (voltage) on the
terminals leading current in and out of that circuit.
If "q is amount of charge moved in time "t:
!PEE !q
P=
=
!V
!t
!t
CJ J
= =W
s C s
Power supplied to any circuit by ideal battery:
!
P = I!V
I
P = I!
"V
Definition of I
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5
Power Dissipated by a Resistor
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Power Supplied by a Real Battery
P = I!V
I
"V
Ideal battery
P = !I
From definition of resistance:
P=
6
!V 2
R
I
!"
P=I R
2
any
circuit
"V
Real battery
P = ! I " I 2 rint
get real
I
rint
any
circuit
"V
!"
Loop rule: ! " Irint " #V = 0
!V = " # Irint
P = !VI = " I # I 2 rint
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8
RC Circuits
Using Calculus....
•! Charging a capacitor:
C initially uncharged; connect
switch to a at t=0
•! For RC circuit, the voltage on
capacitor changes in time:
Calculate current and charge
as function of time.
Vc =
•! Apply Kirchhoff’s Voltage Law:
IR
Capacitive Time Constant: #
Q/C
! = RC
! - "VR - "VC = 0
Units of ! :
! - IR - QC = 0
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!F =
9
Charging a Capacitor
I=
VC
C
=
=s
A V C/s
The greater the ! "# $, the
greater the charging time.
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Charging a Capacitor
Q = C! (1 " e"t /# )
at!
Q
= ! (1 " e"t /# )
C
10
DEMO
t=0
t=!
t ="
! "t /#
e
R
at! t = 0
t=!
t ="
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11
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12
RC Circuits
Discharging a Capacitor
•! Discharging a capacitor:
Q = C! e
•! C initially charged with
Q=C!!"
t
RC
at! t = 0
t=!
t ="
•! Connect switch S2 at t=0.
•! For RC circuit, the voltage on
capacitor changes in time:
Vc =
"
I=
Q
= ! e"t /#
C
!" ! RCt
e
R
at! t = 0
t=!
t ="
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13
Combinations of RC
DEMO
14
Behavior of Capacitors
R
R
!
C
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•! Charging
–! Initially, the capacitor behaves like a wire.
–! After a long time, the capacitor behaves like an open
switch in terms of current flow.
&C #
' = ( 2 R )$ ! = RC
%2"
•! Discharging
–! Initially, the capacitor behaves like a variable battery.
–! After a long time, the capacitor behaves like an open
switch
C
R
!"
&R#
' = ( 2C)$ ! = RC
%2"
R
C
C
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15
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16
Example
•! At some time after the switch is closed, how much energy is stored
in C if I=2.0 mA. Assume ! = 50V, R = 5K" and C = 40µF
VR
"#
R
C
VC
S
•! Use the current I to find
VR = IR = 2 " 10 #3 A " 5 " 10 3 ! = 10V
•! Use Kirchhoff’s Voltage Rule
VC = " ! VR = 50V ! 10V = 40V
•! The energy stored in the capacitor C is:
U=
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1
1
CVC2 = 40 ! 10 " 6 F ! (40V ) 2
2
2
U = 32mJ
17