Download Physics 202, Lecture 10 Exam 1 Result Basic Circuit Components

Exam 1 Result
Physics 202, Lecture 10
Today’s Topics
Phy202 Exam1 Scores
Direct Current Circuits (Ch. 28)
60
„ Basic circuit components( ε, R, …)
50
„ Kirchhoff’s Rules
40
„ Circuits Analysis (For circuits of R’s and ε’s)
Frequen
ncy
„
Median: 69
A: [83-97]
AB,B: [69-82]
BC,C: [45-68]
D: [<45]
30
20
„ Preview Requirements:
emf, junction rule, loop rule, ….
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Bin
™ Next Tuesday:
RC circuit (ch. 28.4), Magnetic Field(ch. 29.1-3)
Basic Circuit Components
Component
Symbol
Behavior in circuit
ΔV=V+-V- =ε
Ideal battery, emf
Resistor
Realistic Battery
(Ideal) wire
Capacitor
Æ
ΔV= -IR
ε r
ΔV=0 (ÆR=0, L=0, C=0)
ΔV=V- - V+ = - q/C, dq/dt =I
Inductor
ΔV= - LdI/dt
(Ideal) Switch
L=0, C=0, R=0 (on), R=∞ (off)
Note: Letter boundaries are for reference only. ( Based on nominal
A=15%, B=50% etc. and subject to final curve at the end of semester)
Simple Circuit 1: Resistors In Series
‰ Exercise: show Req=R1+ R2.
ƒ I1=I2=I
ƒ ΔV =VR1+VR2
= IR1+IR2
Æ ΔV =I(R1+R2)
i Req=R
i.e.
R1+R2
R1
R2
I1
I2
I
R1 +R2
Transformer
Diodes,
Transistors,…
Future Topics
‰ In general: Req=R1+ R2+ R3+ ….
1
Quiz/Exercise: Equivalent Resistance of a
Combined Parallel and Serial Circuit
Simple Circuit 2: Resistors In Parallel
VR1=I1R1
‰ Show 1/Req=1/R1+ 1/R2.
ƒ VR1=VR2 =ΔV
ƒ I1 + I2= I
ÆI1R1 /R1 + I2R2 /R2=I
ÆΔV (1/R1+1/R2)=I
ÆΔV=I 1/ ((1/R1+1/R2)
i.e.
‰ What is the Req for the combination shown?
R1= R2=1Ω, R3=2Ω, R4=4Ω.
I1
VR2=I2R2 I
2
I
Req
R1
1.
2.
3
Î 3.
4.
8Ω
6Ω
5Ω
None of above
R1
R2
R4
R3
1/(1/R1 +1/R2)
R2
‰ In general: 1/Req=1/R1+ 1/R2+ 1/R3+ ….
A Complicated Circuit
loops
Kirchhoff’s Rules: Junction Rule
‰ Junction Rule (Charge conservation):
I1=I2+I3
The sum of currents entering any
junction equals the sum of currents
leaving that junction.
A complicated circuit :
ƒMay contain more than one emf
ƒMay not be simplified as
“in series” or “in parallel”
ƒ May contain multi loops and
junctions.
ΣIin =ΣIout
‰ In practice, the classifications of “in” and “out” are determined by
assigned direction for each current.
The assignment of current directions can be arbitrary. They may not
be the same as actual directions, which are not known a prior.
ƒ “in” : current with assigned direction towards junction
ƒ “out” : current with assigned direction off junction
in
in
junctions
out
2
(Very) Quick Quiz: Junction Rule
Quick Quiz: Junction Rule
‰ What is the junction rule for the current assignment
shown?
1. I1+I2=I3
I1
2. I1=I2+I3 Í
I2
I3
3. I1-I2=I3
‰ What is the junction rule for the current assignment
shown?
1. I1+I2=I3
I1
2. I1+I2+I3 =0 Í
I2
I3
3. Neither
Although equation 2 and 3 are equivalent,
equation 3 does not follow template form Iin=Iout
While the actual currents can not all goes into a junction,
the assigned currents can.
Kirchhoff’s Rules: Loop Rule
‰ Loop Rule (Energy Conservation):
The sum of potential drops across
components along any closed
circuit loop must be zero.
Σ ΔV =0
¾ The potential “drop” across a component is always defined as
Vdown_stream_end –V up_sream_end where,
the stream direction is the same as loop direction
¾ The exact expression of the potential drop is determined by the
type of component and the assigned current direction. (See
next slides)
Determine Potential Difference
b
a
b
a
loop
direction
loop
direction
ΔV=Vb-Va= - ε
ΔV=Vb-Va= +ε
I
I
b
a
loop
direction
ΔV=Vb-Va= -IR
b
a
loop
direction
ΔV=Vb-Va= +IR
3
Steps to Apply Kirchhoff’s Rules
Example 1: Multi-Loop
1. Assign a directional current for each branch (segment) of a
circuit. The assigned direction for each current can be arbitrarily
chosen but, once assigned, need to be observed.
2. Set up junction rules at certain (any) junctions.
Normally, # of junctions = # of branches -1.
3 Select a number of closed loops to apply loop rule.
3.
rule
For each closed loop, assign a loop direction (clockwise or
counter clockwise). Follow that assigned direction, find ΔV drop
across each component, and apply loop rule.
# of loops determined by # of unknowns.
4. Solve for unknowns.
5. If a current is found to be negative, it means its actual direction
is opposite to the originally chosen one. The magnitude is
always correct.
3.0A
2.0A
1.0A
Actual situation
ε2
ε1
R1
R3
Example 1 Again: Different Initial Directions
Example 1: Interpretation of Results
I1= 2.0A, I2= -3.0A, I3= -1.0A,
‰ (Text example 28.9)
Find out I1, I2, I3
‰ Kirchhoff’s Rules:
Junction c:
R2
I1+I2=I3
L
Loop
abcda:
b d
ε1 - I1R1 - I3R3 =0
Loop befcb:
-ε2 + I1R1 –ε1 - I2R2 =0
Solving three equations:
I1= 2.0A, I2= -3.0A, I3= -1.0A,
What does the – sign mean?
Different initial direction for I1, I2
‰ Apply Kirchhoff’s Rules:
Junction c:
0=I3+I1+I2
R2
Loop abcda:
ε1
ε1 + I1R1 - I3R3 =0
0
Loop befcb:
-ε2 - I1R1 -ε1 + I2R2 =0
Solving three equations:
I1= -2.0A, I2= +3.0A, I3= -1.0A,
ε2
R1
R3
Same effective result as in previous slide
4