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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