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ELECTRICITY TOPIC #1 IMPORTANT TERMS TO KNOW. 1. breadboard: circuit board on which electrical components are attached 2. conducting: (material that)

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Presentation on theme: "ELECTRICITY TOPIC #1 IMPORTANT TERMS TO KNOW. 1. breadboard: circuit board on which electrical components are attached 2. conducting: (material that)"— Presentation transcript:

1 ELECTRICITY TOPIC #1 IMPORTANT TERMS TO KNOW

2 1. breadboard: circuit board on which electrical components are attached 2. conducting: (material that) serves as a channel for electricity 3. terminals: point where electiric current enters or leaves

3 4. insulation: prevents/restricts flow of electricity 5. electricity: flow of electrons 6. series: one end to end connection of electrical components

4 7. Current: flow of electricity 8. battery: storage of electricity 9. filament: thread-like conductor that lights up when current flows through it

5 10. circuit: complete path of an electrical current 11. parallel: two electrical components in the same path

6 What happens to the voltage when you connect a set of batteries together in a series?

7 The voltage of the batteries combines increasing the total voltage 1.5 volts +1.5 volts 3.0 volts

8 The figure, below, illustrates a battery pack in which “cell 3” produces only 0.6V instead of the full 1.20V. With depressed operating voltage, this battery reaches the end-of-discharge point sooner than a normal pack. The voltage collapses and the device turns off with “Low Battery” message.

9 The figure, below, illustrates four cells connected in parallel. The voltage of the pack of 4 batteries remains at 1.20V, but the current handling and runtime are increased fourfold because when batteries are connected in parallel the current gets divided between the two batteries and so the runtime will increase.

10 A weak cell will not affect the voltage but will provide a low runtime due to reduced current handling. A shorted cell could cause excessive heat and become a fire hazard.

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12 ELECTRICITY TOPIC #2: Identifying Resistors and using Ohm’s Law

13 I = E/R “I” stands for

14 I = E/R “I” stands for current

15 I = E/R “I” stands for current “E” stands for

16 I = E/R “I” stands for current “E” stands for volts

17 I = E/R “I” stands for current “E” stands for volts “R” stands for

18 I = E/R “I” stands for current “E” stands for volts “R” stands for resistance

19 The ampere (symbol: A) is the SI unit of electric current [1] (symbol: I) and is one of the seven [2] SI base units. It is named after André- Marie Ampère (1775–1836), French mathematician and physicist, considered the father of electrodynamics. In practice, its name is often shortened to amp.SI electric current [1] [2]SI base unitsAndré- Marie Ampère In practical terms, the ampere is a measure of the amount of electric charge passing a point per unit time. Around 6.241 × 10 18 electrons passing a given point each second constitutes one ampere. [3]electric chargeelectrons [3]

20 What is a volt, by the way? A volt is the international scientific term used to describe the electric force of a battery (a.k.a. the electromotive force). Volts are determined by finding the potential energy that would be released when one ampere of current is positioned against one ohm of resistance.

21 ELECTRICITY TOPIC #3: Resistors in Series and Parallel circuits

22 In a series circuit you would add the total resistance and divide it into the total voltage

23 Series Circuit With One Resistor

24 Series Circuit With One Resistor 3 amps

25 What if you had more than one resistor, though?

26

27 Series Circuit With Three Resistors

28 Series Circuit With Three Resistors

29 Series Circuit With Three Resistors (0.5 amps)

30 In a parallel circuit you would add the reciprocals of all the resistors and divide that total into the number 1. Then you would take this answer and divide it into the total voltage ????

31 1/10 + 1/2 + 1/1 = 1/10 + 5/10 + 10/10 = 16/10 = 8/5 SO…..

32 1/10 + 1/2 + 1/1 = 1/10 + 5/10 + 10/10 = 16/10 = 8/5 SO….. 1 divided by 8/5 = 1/1 times 5/8 = 5/8 =.625 FINALLY…..

33 1/10 + 1/2 + 1/1 = 1/10 + 5/10 + 10/10 = 16/10 = 8/5 SO….. 1 divided by 8/5 = 1/1 times 5/8 = 5/8 =.625 FINALLY….. 9 divided by.625 = 14.4 amps

34 How about a more simpler way for our sixth grade class?

35 For the sixth grade we’ll use all the same ohm resistors. This is the only condition under which this “trick” will work, though. Let’s say we had a circuit with 50 volts and two 5 ohm resistors.

36 For the sixth grade we’ll use all the same ohm resistors. This is the only condition under which this “trick” will work, though. Let’s say we had a circuit with 50 volts and two 5 ohm resistors. Divide resistor value by the number of resistors in the circuit. Then divide the total voltage by this quotient.

37 For the sixth grade we’ll use all the same ohm resistors. This is the only condition under which this “trick” will work, though. Let’s say we had a circuit with 50 volts and two 5 ohm resistors. Divide resistor value by the number of resistors in the circuit. Then divide the total voltage by this quotient. 5 ohms divided by 2 resistors equals 2.5 50 divided by 2.5 equals 20 amps

38 increased.

39 Electrical Fuse Lab


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