Science Olympiad Shock Value Activities involving basic understanding

Science Olympiad Shock Value

Activities involving basic understanding of electricity, magnetism and simple electrical devices. Hopefully this will involve test type questions and hands -on activities. Could be: • stations. • building specific circuits. • analyzing graphs. • mapping fields. • there are many possibilities.

Series Circuit In a series circuit the Equivalent Resistance is the sum of the individual resistances. R 1 = 4. 0 Ω R 2 = 8. 0 Ω R 3 = 6. 0 Ω RS = R 1 + R 2 + R 3 = 4. 0 + 8. 0 + 6. 0 = 18. 0 Ω A quick check is the Equivalent Resistance for resistors in Series is always more than the largest individual resistance. In the above example, RS must be greater than 8. 0 Ω.

Parallel Circuit In a parallel circuit the Equivalent Resistance is the reciprocal of the sum of the reciprocal of the individual resistances. R 1 = 4. 0 Ω R 2 = 8. 0 Ω R 3 = 6. 0 Ω 1/RP = 1/R 1 + 1/R 2 + 1/R 3 1/RP = 1/4. 0 + 1/8. 0 +1/6. 0 1/RP = 13/24 Ω RP = 24/13 Ω = 1. 85 Ω A quick check is the Equivalent Resistance for resistors in Parallel is always less than the smallest individual resistance. In this case RP must be less than 4. 0 Ω. Using your calculator: RP = (R 1 -1 + R 2 -1 + R 3 -1)-1 = (4 -1 + 8 -1 + 6 -1)-1 = 1. 85 Ω

Using the circuit shown below, what is the current passing through each resistor? R 1 = 5. 40 Ω 30. 0 V R 2 = 6. 20 Ω R 4 = 16. 0 Ω R 3 = 9. 80 Ω R 5 = 8. 00 Ω R 1 = 5. 40 Ω 30. 0 V RS 1 = 16. 0 Ω RS 2 = 24. 0 Ω In analyzing circuits, you can replace any group of resistors with its equivalent. RS 1 = R 2 + R 3 = 6. 20 Ω + 9. 80 Ω = 16. 0 Ω RS 2 = R 4 + R 5 = 16. 0 Ω + 8. 0 Ω = 24. 0 Ω

Using the circuit shown below, what is the current passing through each resistor? R 1 = 5. 40 Ω 30. 0 V R 2 = 6. 20 Ω R 4 = 16. 0 Ω R 3 = 9. 80 Ω R 5 = 8. 00 Ω R 1 = 5. 40 Ω 30. 0 V RP =9. 60 Ω R 1 = 5. 40 Ω 30. 0 V RS 1 = 16. 0 Ω RS 2 = 24. 0 Ω 1/RP = 1/RS 1 + RS 2 1/RP = 1/16. 0 + 1/24. 0 = 5/48. 0 RP = 48. 0/5 = 9. 60 Ω

Using the circuit shown below, what is the current passing through each resistor? R 1 = 5. 40 Ω 30. 0 V R 2 = 6. 20 Ω R 4 = 16. 0 Ω R 3 = 9. 80 Ω R 5 = 8. 00 Ω R 1 = 5. 40 Ω 30. 0 V RP =9. 60 Ω R 1 = 5. 40 Ω 30. 0 V RS 1 = 16. 0 Ω RS 2 = 24. 0 Ω RTS = RP + R 1 RTS = 5. 40 Ω + 9. 60 Ω = 15. 0 Ω IS = I 1 = IP = VS/RTS = (30. 0 V)/(15. 0 Ω) IS = I 1 = IP = 2. 00 A

Using the circuit shown below, what is the current passing through each resistor? R 1 = 5. 40 Ω 30. 0 V R 2 = 6. 20 Ω R 4 = 16. 0 Ω R 3 = 9. 80 Ω R 5 = 8. 00 Ω R 1 = 5. 40 Ω 30. 0 V RP =9. 60 Ω R 1 = 5. 40 Ω 30. 0 V RS 1 = 16. 0 Ω RS 2 = 24. 0 Ω VP = IPRP = (2. 00 A)(9. 60 Ω) = 19. 2 V VS 1 = VS 2 = VP = 19. 2 V I 4 = I 5 = VS 2/RS 2 = 19. 2 V/24. 0 Ω I 4 = I 5 = 0. 80 A

Using the circuit shown below, what is the current passing through each resistor? R 1 = 5. 40 Ω 30. 0 V R 2 = 6. 20 Ω R 4 = 16. 0 Ω R 3 = 9. 80 Ω R 5 = 8. 00 Ω R 1 = 5. 40 Ω 30. 0 V RP =9. 60 Ω R 1 = 5. 40 Ω 30. 0 V RS 1 = 16. 0 Ω RS 2 = 24. 0 Ω VP = IPRP = (2. 00 A)(9. 60 Ω) = 19. 2 V VS 1 = VS 2 = VP = 19. 2 V I 2 = I 3 = VS 1/RS 1 = 19. 2 V/16. 0 Ω I 2 = I 3 = 1. 20 A

Using the circuit shown below, what is the voltage drop across each resistor? R 1 = 5. 40 Ω 30. 0 V R 2 = 6. 20 Ω R 4 = 16. 0 Ω R 3 = 9. 80 Ω R 5 = 8. 00 Ω V 1 = I 1 R 1 = (2. 00 A)(5. 40 Ω) = 10. 8 V V 2 = I 2 R 2 = (1. 20 A)(6. 20 Ω) = 7. 44 V V 3 = I 3 R 3 = (1. 20 A)(9. 80 Ω) = 11. 76 V I 1 = 2. 0 A I 2 = I 3 = 1. 2 A I 4 = I 5 = 0. 8 A V 4 = I 4 R 4 = (0. 8 A)(16. 0 Ω) = 12. 8 V V 5 = I 5 R 15 = (0. 8 A)(8. 00 Ω) = 6. 4 V

Working with circuits. The challenge here is to be able to apply the basic concepts to a setup in the lab. Either an existing one or have you build a given circuit. There are many ways to achieve this. Here are some examples that were used at Nationals.

Before you is a circuit with switches, identical light bulbs and a battery holder. Draw a schematic of the circuit as if the following were true. There are batteries in both holders that were wired properly in series. • There is a voltmeter placed to measure the voltage drop across Bulb #1. • There is an ammeter placed to measure the current passing through just bulb #3.

Before you is a circuit with switches, identical light bulbs and a battery holder. Draw a schematic of the circuit as if the following were true. There are batteries in both holders that were wired properly in series. • There is a voltmeter placed to measure the voltage drop across Bulb #1. • There is an ammeter placed to measure the current passing through just bulb #3. Bulb #2 Switch #1 Bulb #1 A Bulb #3 A V Switch #2

Breadboard or Circuit Board A Breadboard is an easy way of wiring circuits together. It saves twisting wires together or using lots of leads with alligator clips.

Breadboard or Circuit Board All the connections on the “X” row are connected together. All the connections on the “Y” row are connected together.

Breadboard or Circuit Board Columns are in two sections: 1 A-1 E are connected together. Same for the other columns in this section. Columns 1 F-1 J are connected together, but NOT to 1 A-1 E or adjacent columns in this section.

Breadboard or Circuit Board This is a picture of the breadboards that I brought along. There are many different manufacturers, but all have the two long strips on either side and the two sections of columns in between.

Breadboard Used at Nationals There are three types of resistors used on this board. The possibilities are: Brown Green Black Gold Red Black Gold Orange Black Gold Yellow Violet Black Gold Green Blue Black Gold Brown Black Brown Gold Red Brown Gold

The first two bands give the first two digits of the resistance. The third band is the multiplier that gives the power of ten of the resistance value. The fourth band gives the tolerance of the resistor. 1 st & 2 nd Band Black 0 Brown 1 Red 2 Orange 3 Yellow 4 Green 5 Blue 6 Violet 7 Gray 8 White 9 3 rd Band Multiplier Black 0 x 100 Brown 1 x 101 Red 2 x 102 Orange 3 x 103 Yellow 4 x 104 Green 5 x 105 Blue 6 x 106 Silver Gold -2 x 10 -2 -1 x 10 -1 4 th Band Tolerance Gold 5% Silver 10% none 20%

Practice: Orange Violet Red Gold Orange Violet = 37 1 st & 2 nd Band Black 0 Brown 1 Red 2 Orange 3 Yellow 4 Green 5 Blue 6 Violet 7 Gray 8 White 9 Red = x 102 3 rd Band Multiplier Black 0 x 100 Brown 1 x 101 Red 2 x 102 Orange 3 x 103 Yellow 4 x 104 Green 5 x 105 Blue 6 x 106 Silver Gold -2 x 10 -2 -1 x 10 -1 Gold = 5% TOLERANCE 37 x 102 Ω = 3700 Ω 3700 ± 5% 4 th Band Tolerance Gold 5% Silver 10% none 20%

There are three types of resistors used on this board. The possibilities are: Brown Green Black Gold 15 Ω Red Black Gold 22 Ω Orange Black Gold 33 Ω Yellow Violet Black Gold 47 Ω Green Blue Black Gold 56 Ω Brown Black Brown Gold 100 Ω Red Brown Gold 220 Ω

Breadboard Used at Nationals One exposed: Brown Green Black Gold 15 Ω There are three types of resistors used on this board. The possibilities are: Brown Green Black Gold Red Black Gold Orange Black Gold Yellow Violet Black Gold Green Blue Black Gold Brown Black Brown Gold Red Brown Gold

Breadboard Used at Nationals 56 Ω 220 Ω 15 Ω Single Resistors. Check resistance with meter. There are three types of resistors used on this board. The possibilities are: Brown Green Black Gold Red Black Gold Orange Black Gold Yellow Violet Black Gold Green Blue Black Gold Brown Black Brown Gold Red Brown Gold

Breadboard Used at Nationals Now there are some combinations. Check resistance with meter: 55 to 56 Ω Possible Resistors: 15 Ω, 56 Ω, 220 Ω Four resistors in parallel. Can’t be 15 Ω or 56 Ω 1/RP = 1/220 +1/220 = 4/220 RP = 55 Ω Aha, each resistor is 220 Ω

Breadboard Used at Nationals Now there are some combinations. Check resistance with meter: 11 to 12 Ω Possible Resistors: 15 Ω, 56 Ω, 220 Ω Two resistors in parallel. One is 15 Ω and the other 56 Ω? RP = (15 -1 + 220 -1)-1 =14. 0 Ω too big RP = (15 -1 + 56 -1)-1 = 11. 8 Ω No way to tell which is which.

Breadboard Used at Nationals The nasty one Check resistance with meter: 54 - 55 Ω R 6 = 15 Ω R 7 = ? Ω R 8 = can’t be 15 Ω or 56 Ω so that leaves 220 Ω

Breadboard Used at Nationals The nasty one Check resistance with meter: 54 -55 Ω R 6 = 15 Ω R 7 = ? Ω R 8 = 220 Ω RS = 15 + 56 =71 Ω RP = (71 -1 + 220 -1)-1 = 54 Ω

Concerns: • Less expensive meters seem to be less accurate. • When students experiment using multimeters, they often wire an ammeter in parallel with a device creating a short and thus a large current which may blow the fuse in the meter. You may not have received replacement fuses and if you have them, they are not easy to replace (you will need a screw driver). • If you have many teams competing at the same time, there is a need for many setups. • Colleges and Universities have sophisticated equipment. There may be a need to explain how they function.