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ANNOUNCEMENT • Exam 1: Tuesday September 27, 2016, 8 PM - 10 PM • Location: Elliot Hall of Music • Covers all readings, lectures, homework from
Chapters 21 through 23 – Multiple choice (15-18 questions)
• Practice exams – On the course website and on CHIP
• Bring your student ID card and your own one-page (two-sides) crib sheet – Only a few equations will be given
• The equation sheet that will be given with the exam is posted on the course homepage – Link on the right labeled “Equationsheet”
– It is your responsibility to create your own crib sheet
ANNOUNCEMENT • Crib sheet
– 8.5 x 11 – Can be handwritten, computer-generated, painted, etc.
• All calculators allowed – Except web-enabled, internet or bluetooth connected
• Nothing that can communicate with other devices
http://www.wiseguysynth.com/larry/kw2010/2010_construction.htm
Resistors and Circuits
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V
R1 R2
R4
R3
Reffective =R1+R2 + ...
1Reffective
=1R1
+1
R2
+ ...
This Class • Resistors in series:
– Current through is same
– Voltage drop across is Iri
• Resistors in parallel: – Voltage drop across is same
– Current through is V/Ri
• Solve Circuits
R1 R2
R1
R2
Resistors in Series
• Voltage drop across resistors in series – Each has identical current
• Can replace with equivalent resistor – Same total potential drop, same current
Va −Vb = IR1 Vb −Vc = IR2
Vtotal =Vab +Vbc = IR1+ IR2 = I(R1+R2)
Req =R1+R2
Another (Intuitive) Way
• Consider two cylindrical resistors with lengths L1 and L2
• Put them together, end to end to make a longer one...
V
R1
R2
L2
L1
R1 = ρ
L1
A
Req = ρ
L1+L2
A=R1+R2 R =R1+R2
R2 = ρ
L2
A
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Resistors in Parallel
• Voltage drop identical across resistors in parallel – Current can vary across each
• Can replace with equivalent resistor – Same potential drop, same total
current
V = I1R1 = I2R2
I1 =
VR1
I2 =VR2
Itotal =
VReq
=VR1
+VR2
1Req
=1R1
+1
R2
Another (Intuitive) Way
• Consider two cylindrical resistors of equal length L with cross-sectional areas A1 and A2
• Put them together, side by side … to make one “fatter”one
R1 = ρ
LA1
Req =
ρLA1+A2( )
⇒
1Req
=A1
ρL+
A2
ρL
=1R1
+1
R2
V
A1 A2
R2 = ρ
LA2
1Req
=1R1
+1
R2
R1 R2 L L
Example 1 • Consider the ideal circuit shown:
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)
12V I1 I2
a b
d c
50Ω
20Ω 80Ω
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Example 1 • Consider the ideal circuit shown:
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)
12V I1 I2
a b
d c
50Ω
20Ω 80Ω
• Assume cd is a perfect conductor
• Still an equipotential even though this example is not static
=> Points d and c are the same, electrically
Example 2
12V I1 I2
a b
d c
50Ω
20Ω 80Ω
(a) I1 > I2 (b) I1 = I2 (c) I1 < I2
What is the relationship between I1 and I2?
• Consider the ideal circuit shown:
Example 2
12V I1 I2
a b
d c
50Ω
20Ω 80Ω
(a) I1 > I2 (b) I1 = I2 (c) I1 < I2
What is the relationship between I1 and I2?
• Consider the ideal circuit shown:
• Vb -Vd = Vb -Vc assuming perfectly conducting wires • Therefore,
I1(20Ω) = I2(80Ω) I1 = 4I2
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Kirchhoff’s First Rule (“Loop Rule” or “Kirchhoff’s Voltage Law”)
• The algebraic sum of the changes in potential in a complete traversal of any loop of circuit must be zero – Based on energy conservation
– A restatement that the potential difference is independent of path
• Applies to any circuit
Move around circuit:
KVL : Vn = 0
loop∑ε1
R1 R2 I
ε2
- + + -
+ ε1 - IR1 - IR2 - ε2 = 0
Rules
• Loop direction is ARBITRARY
• Voltage gains enter equation with a + sign
• Voltage drops enter equation with a - sign
• Battery traverse
– - terminal to + terminal: V increases => +ε
– + terminal to - terminal: V drops => -ε
ε1
R1 R2 I
ε2
- + + -
+ ε1 - IR1 - IR2 - ε2 = 0
Rules
• Resistor traverse – Positive direction
• Voltage drops – Enters the equation with a – sign (-IR)
– Negative direction
• Voltage rises – Enters the equation with a + sign (+IR)
Note: e always points from negative to positve
ε1
R1 R2 I
ε2
- + + -
+ ε1 - IR1 - IR2 - ε2 = 0
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Rules
• No “wrong” set of paths (multi-loop circuits)
• Flip one path, just change every sign in that path’s equation
• BUT – One or more of the currents in your solution may be NEGATIVE
– Means actual current flow is opposite to the path you chose
• Where ONLY that one current is flowing
ε1
R1 R2 I
ε2
- + + -
+ ε1 - IR1 - IR2 - ε2 = 0
Loop Example
a
d
b
e c
f
R1
I
R2 R3
R4
I
ε1
ε2
+ -
+ -
Vn = 0
loop∑
I =
ε1 −ε2
R1+R2 +R3 +R4
If ε1 < ε2 , I will be negative, i.e., will flow clockwise, opposite to path
⇒ − IR1 − IR2 −ε2 − IR3 − iR4 +ε1 = 0
Traverse loop from a to f in this case
Internal Resistance of an EMF Device
• Any real emf device has internal resistance – e.g., a real battery
• Apply Kirchhoff’s Loop Rule (clockwise)
ε − Ir − IR = 0→ I = ε
R + r Vab = ε − Ir = ε R
R + r
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Kirchhoff’s Second Rule (Junction Rule or “Kirchhoff’s Current Law”)
• The sum of the current entering any junction (or ‘node’) must equal the sum of the currents leaving that junction
– Conservation of charge
• Branch currents – Currents entering and leaving circuit nodes – Each distinct branch must be assigned a current, Ii
I I1
I2 Iin = Iout∑
How to Use Kirchhoff’s Laws
• Analyze the circuit & identify all circuit nodes – Use KCL
1) I1 = I2 + I3
• Identify all independent loops – Use KVL
2) –ε1 – I3R3 – I1R1 = 0 3) –ε1 + ε2 – I2R2 – I1R1 = 0 4) = (3)-(2) = ε2 – I2R2 + I3R3 = 0 Only 2 are independent
ε1
+ -
ε2
- +
R3 R2
R1
I1
I2
I3
a b c
d
How to Use Kirchhoff’s Laws
• Solve for I1, I2, and I3 – Find I2 and I3 in terms of I1
– Solve for I1 from eqn. (1)
ε1
+ -
ε2
- +
R3 R2
R1
I1
I2
I3
a b c
d
From eqn. (2)
I3 = − ε1+ I1R1( ) / R3
I2 = ε2 −ε1 − I1R1( ) / R2
From eqn. (3)
I1 =
ε2 −ε1
R2
−ε1
R3
− I1(R1
R2
+R1
R3
)
I1 =
ε2 −ε1
R2
−ε1
R3
1+ R1
R2
+R1
R3
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How to Use Kirchhoff’s Laws
• Only works here since only 2 currents in Eqns. (2) & (3)
• More general case – 3 simultaneous eqns. – Can use standard software
ε1
+ -
ε2
- +
R3 R2
R1
I1
I2
I3
a b c
d
Example 25-16
I = I1+ I2
Loop 1 (abcdefa): −2I2 −5−3(I1+ I2)+12 = 0
7−3I1 −5I2 = 0
Loop 2 (bcdeb): −2I2 −5+ 4I1 = 0
2 equations in 2 unknowns
−3I1 −5I2 = −7⎡⎣ ⎤⎦(2)
4I1 − 2I2 = 5⎡⎣ ⎤⎦(−5)
−6I1 −10I2 = −14−20I1+10I2 = −25
I1 =1.5 A
−3 1.5( )−5I2 = −7
I2 = 0.5 A
Cramer’s Rule
If
a1x+b1y = c1
a2x+b2y = c2
x =
c1 b1
c2 b2
a1 b1
a2 b2
=c1b2 −c2b1
a1b2 −a2b1
y =
a1 c1
a2 c2
a1 b1
a2 b2
=a1c2 −a2c1
a1b2 −a2b1
Then
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Summary of Resistor & Capacitor Combinations
Req = Rii=1
n
∑
1Req
=1Rii=1
n
∑
Req =R1R2
R1+R2
for n = 2
Resistors Capacitors
Series
Parallel
1Ceq
=1Cii=1
n
∑
Ceq =C1C2
C1+C2
for n = 2
Ceq = Cii=1
n
∑
Summary of Simple Circuits
• Resistors in series: – Current through is same – Voltage drop across is IRi
• Resistors in parallel: – Voltage drop across is same – Current through is V /Ri
Req =R1+R2 +R3 + ...
1Req
=1R1
+1
R2
+1
R3
+ ...
Problem Solving Tips
• When you are given a circuit, first carefully analyze circuit topology
– Find the nodes and distinct branches
• Pick Linearly Independent subsets of each
– Assign branch currents
• Use Kirchhoff’s First Rule for all independent loops in the circuit
– Sum of the voltages around these loops is zero
• Use Kirchhoff’s Second Rule for all independent nodes in circuit
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Ammeter, A, inserted into the circuit
Voltmeter, V, across a circuit element
I
I
Ammeter & Voltmeter
I
I
Ammeter & Voltmeter • Both disturb or perturb
a circuit • To minimize the
perturbation: – Resistance of A small
compared to R1 + R2 + r • Ideal ammeter has r = 0
– Resistance of V large compared to R1
• Ideal voltmeter has infinite R
Ig
Rg
Ig ~ 50−100 µA for full scale deflection
Rg ~ 100Ω
=
Galvanometer
Scale reading proportional to I
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Vg =RgIg = IPRP
RP =Rg
Ig
IP
=Rg
Ig
I − Ig
If Rg = 100 Ω and Ig = 50 µA, then
RP =100Ω 50×10−6 A
1A−50×10−6 A≅100Ω 50×10−6 A
1A= 0.005Ω
To change the scale, change RP
small parallel resistor (shunt)
Make a 1.0 A Full-Scale Ammeter
Ig(RS +Rg ) =V
RS +Rg =100V
50×10−6 A= 2×106Ω
Assume Rg = 100 Ω and Ig = 50 µA for full scale deflection (typical). Make a 100 V full scale deflection voltmeter.
100 Ω negligible
Voltmeter
Ohmmeter Battery in series with galvanometer and resistor
• Full-scale when a-b is shorted • Reads resistance when resistor connected across a-b
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