Chapter 27 Kirchoffs Rules

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Chapter 27 Kirchoffs Rules

• PHYS 2326-14

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Concepts to Know

• ElectroMotiveForce EMF
• Junction
• Loop
• dArsonval Galvanometer
• Ammeter
• Voltmeter
• Ohmmeter
• Potentiometer
• Multimeter

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Electromotive Force - EMF

• e electromotive force a force that creates a
  charge such as the chemical reaction in a battery
  which builds up a voltage potential.
• ?V e Ir where I is the current and r is the
  internal resistance of the battery or device.
• I e/(R r)

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Load Matching

• See example 28.2
• Net result maximum power is transferred when the
  load resistance R r internal resistance.

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Ohms Law Review

• I V/R, V IR, R V/I
• Power Watts Joules /sec
• P IV I2 R V2/R
• Series Resistances add
• Parallel Resistances 1/Rt 1/R1 1/R2
• opposite from capacitors

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Kirchoffs Rules

• Junction Rule at any junction (3 or more
  connections) the sum of all currents 0
• there can be no charge buildup at the junction
• Loop Rule The sum of the potential differences
  across all elements around a closed loop must 0

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Potential Difference across a component

• Going around the loop mark a plus by the end of a
  component on the side where the assigned current
  is coming from. Going with the current flow a
  resistor drops voltage a battery increases going
  to which is going downstream. Going against
  the assigned current flow a resistor increases V
  and a battery decreases from to -.

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Example 1 Kirchoffs Rules

• R1 1O, R2 1 O, R3 3 O, R4 4 O,
• R5 5 O, R6 6 O

R4
R2
b
d
a
R3
10V

R1
R5

5V
c
R6
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Example 1

• Assign currents for the loops
• Determine resistor high V sides for these
  currents and place signs

I2
10V
I1




I3

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Example 1

• Set up equations for the 3 loops
• -1O I1 - 2 O I2 - 3O I1 3O I2 5V 0
• 3 O I1 - 3 O I2 - 4 O I2 10V - 5 O I2 5 O I3
• -5V 5 O I2 - 5 O I3 - 6 O I3
• Rearrange
• 6 I1 3I2 0 I3 5 V/ O 5 A
• -3 I1 12 I2 5 I3 10 A
• 0 I1 5 I2 11 I3 -5A

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Solve by determinants

• Make determinant for equations, denominator first
  then solve for each I column

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Solving for Currents
13

• Repeating for the final current I3

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dArsonval galvanometer

• Standard analog meter, typical might have 10 ohms
  and full scale current of 1ma
• If low resistance shunt across it becomes an
  ammeter
• If resistor added in series reads voltage

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Example 2

• Given dArsonval galvanometer with 10 Ohms and
  full scale reading at 1ma
• find a) min. voltage for F.S., b)shunt needed for
  ammeter F.S. 0.1A, c) Series resistance needed
  for voltmeter at 20V F.S. and d) what voltage is
  read at one-quarter full scale

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Example 2

• Rg 10 O, current at F.S. 0.001 A
• Va IR 0.001 10 0.010V or 10mV
• for shunt in parallel with meter R and total
  current Ib, Va Ib Rb, where 1/Rb 1/Rs 1/Rg,
  Rb0.1 O, Rs 0.10101 O
• Vc IgRc, Rc RgRs, Rc 20K O
• Rs 19.99k O
• d) ¼ full scale 20/4 5V

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RC Circuits

• I Io exp(-t/RC) RC time constant tau
• I dq/dt
• I e/R
• See section 28.4
• See example 28.10 for energy in capacitor at time
  t is UQ2/(2C)exp(-2t/RC) falls off twice as
  fast

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Electric Safety

• Basic power for homes comes in 110 VAC and 220
  VAC. There are 3 prong outlets, hot neutral and
  safety ground
• Black is hot, white is neutral, green is safety
  ground. Also the larger spade on the plug is
  neutral. Neutral can be higher than ground due
  to internal IR drops as it carries the return
  current. Green should carry no current.
• ALWAYS Verify in case of faults or wiring errors.
  Beware using test equipment to measure this
  stuff and never use an instrument with a 3rd
  prong ground to measure these voltages.

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Example 3

• Combine Series R1 R2 toR5
• Combine Parallel 1/R5 1/R4 1/R6
• Combine Series R6 R3 R7

R2
R6
c
R1
a
Ro
R4
Ro
R3
R5
R3
b
R4
Ro
R7
R3
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Example 3

• Given Ro 1 O, R1 1O, R2 2 O, R3 3 O
• R4 6 O, Vo 12V Find a) voltage across and
  current through each resistor, b) voltage Vab
  across the battery, c) the electric potential at
  point c.
• Find equivalent resistance and work backwards to
  get voltage and current

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Example 3

• R5 R1R2 12 3 O
• R4 R5 parallel 1/R6 1/R4 1/R5 2 O
• Ro, R6 R3 in series, R7 RoR6R3 6 O
• V IR7, IV/R 2A
• Io I3 I6 I
• Vo IoRo 2 V
• V3 I3 R3 6V
• V6 I6 R6 4V
• V5 V6 I5 R5, I5 1.333A
• V4 V6 I4 R4, I4 0.667A
• I2 I5, V2I2 R2, V2 2.667V

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Example 3

• a) Vo 2 V Io 2 A
• V1 1.333 V I1 1.333 A
• V2 2.667 V I2 1.333 A
• V3 6 V I3 2 A
• V4 4V I4 0.667A
• b) Vab V Vo 10V
• c) Vc Vcb V2V3 8.667V