1 / 44

# unit three: ohm s law - PowerPoint PPT Presentation

ET115 DC Electronics. Unit Three: Ohm’s Law. John Elberfeld [email protected] WWW.J-Elberfeld.com. Schedule. Unit Topic Chpt Labs Quantities, Units, Safety 1 2 (13) Voltage, Current, Resistance 2 3 + 16 Ohm’s Law 3 5 (35) Energy and Power 3 6 (41)

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'unit three: ohm s law' - Jimmy

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Unit Three:Ohm’s Law

John Elberfeld

WWW.J-Elberfeld.com

Unit Topic Chpt Labs

• Quantities, Units, Safety 1 2 (13)

• Voltage, Current, Resistance 2 3 + 16

• Ohm’s Law 3 5 (35)

• Energy and Power 3 6 (41)

• Series Circuits Exam I 4 7 (49)

• Parallel Circuits 5 9 (65)

• Series-Parallel Circuits 6 10 (75)

• Thevenin’s, Power Exam 2 6 19 (133)

• Superposition Theorem 6 11 (81)

• Magnetism & Magnetic Devices 7 Lab Final

• Course Review and Final Exam

• Describe the relationship among voltage, current, and resistance.

• Given two of the three variables in Ohm’s Law, solve for the remaining quantity.

• Solve Ohm’s Law problems using metric prefixes.

• Construct basic DC circuits on a protoboard.

• Use a digital multimeter (DMM) to measure a predetermined low voltage on a power supply.

• Measure resistances and voltages in a DC circuit using a DMM.

• Explain the Multisim workbench and show how to construct a basic circuit.

• Test circuits by connecting simulated instruments in Multisim

• Read and study

• Chapter 3: Ohm’s Law Pages 71-80

• Lab Experiment 5:

• Ohm’s Law Pages 35-38

• Complete all measurements, graphs, and questions and turn in your lab before leaving the room

• Answer all questions on the homework handout

• Be prepared for a quiz on questions similar to those on the homework.

• If there are any calculations, you must show ALL your work for credit:

• Write down the formula

• Show numbers in the formula

• Circle answer with the proper units

• MEMORIZE: V = I R

• Ohm’s Law

• If you increase the voltage, you increase the current proportionally

• 3 times the voltage gives you three times the current

• Resistance (ohms) is the proportionality constant and depends on the atomic structure of the material conducting the current

V

Voltage

x

x

x

x

x

I – Current in Amps

Reasoning

• Ohms Law: V = I R

• High voltage produces high current for a given resistance

• Low voltage produces low current for a given resistance

• For a given voltage, a high resistance produces a low current

• For a given voltage, a low resistance produces a high current

• A battery with the voltage V pushes a current I through a resistor R

V = I R

Ohm’s Law

• This is the BIG IDEA for the day (year)!

• V = I R

• What if we divide both sides by R?

• V = I R R R

• But R/R = 1, so we don’t need to write it down:

• I = V I = V / R R

Ohm’s Law

• V = I R

• What if we divide both sides by I?

• V = I R I I

• But I / I = 1, so we don’t need to write it down:

• R = V R = V / I I

• Memorize: V = I R

• Use algebra to find:

• I = V / R

• R = V / I

• If you can, learn all three variations, but you can get by if you memorize:

V = I R

• V = I R

• What voltage (V) is needed to push a current of 2 Amperes (I) through a resistance of 18 Ohms (R) ?

• V = I R

• What voltage (V) is needed to push a current of 2 Amperes (I) through a resistance of 18 Ohms (R) ?

• V = I R

• V = 2 A x 18 Ω

• V = 36 V

1.2k Ω

575 μA

Examples

• Ohms Law: V = I R k = 103μ = 10-6

• How much voltage must be connected across a 1.2 k Ω resistor to cause 575 μA of current to flow?

• V = I R

Examples

Ohms Law: V = I R k = 103μ = 10-6

How much voltage must be connected across a 1.2 k Ω resistor to cause 575 μA of current to flow?

V = I R

V = 575 μA 1.2 k Ω

V = .69V = 690 x 10-3V = 690 mV

? V

1.2k Ω

575 μA

25 Ω

103=k10-3 = m10-6 = μ

Examples

• Ohms Law: V = I R

• How much current flow through a 25 Ω resistor with 10 V across it?

• V = I R I = V / R

Examples

Ohms Law: V = I R

How much current flow through a 25 Ω resistor with 10 V across it?

V = I R I = V / R

10 V = I 25 Ω

I = 10 V / 25 Ω

I = .4 A or 400 x 10-3A = 400 mA

10 V

25 Ω

103=k10-3 = m10-6 = μ

250 mA

103=k10-3 = m10-6 = μ

Examples

• Ohms Law: V = I R

• If a certain resistor allows 250 mA to flow when 35 V are across it, what is the resistance?

• V = I R R = V / I

250 mA

103=k10-3 = m10-6 = μ

Examples

• Ohms Law: V = I R

• If a certain resistor allows 250 mA to flow when 35 V are across it, what is the resistance?

• V = I R R = V / I

• 35 V = 250 mA R

• R = 35 V / 250 ma

• R = 140 Ω

3.3k Ω

103=k10-3 = m10-6 = μ

Examples

• Ohms Law: V = I R

• How much current flow through a 3.3k Ω resistor with 4.5 mV across it?

• V = I R I = V / R

3.3k Ω

103=k10-3 = m10-6 = μ

Examples

• Ohms Law: V = I R

• How much current flow through a 3.3k Ω resistor with 4.5 mV across it?

• V = I R I = V / R

• 4.5 mV = I 3.3k Ω

• I = 4.5 mV / 3.3k Ω

• I = 1.36 μ A

• V = I R

• What current (I) flows through a resistance of 8 ohms when the resistor is connect to a 24 volt battery?

Practice

• What current (I) flows through a resistance of 8 ohms when the resistor is connect to a 24 volt battery?

• V = I R I = V / R

• 24 V = I x 8 Ω I = 24 V / 8 Ω

• I = 24 V / 8 Ω I = 3 A

• I = 3 A

Practice

• What size resistor allows 2 amperes of current through it when it is connected to a 10 Volt power supply?

Practice

• What size resistor allows 2 amperes of current through it when it is connected to a 10 Volt power supply?

• V = I R R = V / I

• 10 V = 2 A x R R = 10 V / 2 A

• R = 10 V / 2 A R = 5 Ω

• R = 5 Ω

• Ohm’s Law describes the relationship among voltage, current, and resistance – it does not control it!

• In lab, you will prove to yourself that Ohm’s Law applies to circuits

• Use the special handout to organize your information

• Your resistors can off by +/- 5% from the marked value

• You must measure as accurately as possible the real resistance used in your experiment

• Use TWO DMMs in your experiment

• Record as many digits as possible for both voltage and current

• You must BREAK the circuit to measure current

A

V

• Your lab handout says to plot I along the x axis and V along the y axis

• The slope is Δy / Δx = ΔV/ ΔI

• Based on Ohm’s Law, R = V / I, just like the slope

1. Select the correct voltage mode (ac or dc).

2. Select range higher than expected voltage.

3. Connect the meter across the points. Red, positive (+), Black, common (–)

• 4. Reduce the range setting until the reading fails

• 5. Increase the range setting one step and record all the numbers, with the proper units, shown on the meter

• 34.67 mV, for example

• Voltage is always the difference between TWO points.

• Measure VBC by attaching the RED lead to B and the BLACK lead to C

A

B

V

D

C

• If only one letter is given, attach the RED lead to that letter, and the BLACK lead to the reference point or ground.

• If D is your reference point, VB is:

A

B

D

C

V

• If D is your reference point, then

• VB is really VBD

• VC is really VCD

• Electrically, then

• VBC = VBD - VCD

• Voltage is the difference between two points

• Choosing a different reference point does NOT change the real voltage

1. Ohm’s Law

2. Solving for voltage, current, or resistance in a one-load circuit

3. Ohm’s Law using metric prefixes