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Aim: How can we explain a series circuit?

Aim: How can we explain a series circuit?. Do Now: Find the resistance of a 0.25 m Tungsten wire with a diameter of 1.5 x 10 -3 m. R = ρ L A R = (5.6 x 10 -8 Ω ·m) (0.25 m) 1.77 x 10 -6 m 2 R = 7.9 x 10 -3 Ω. Circuits.

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Aim: How can we explain a series circuit?

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  1. Aim: How can we explain a series circuit? Do Now: Find the resistance of a 0.25 m Tungsten wire with a diameter of 1.5 x 10-3 m. R = ρL A R = (5.6 x 10-8Ω·m) (0.25 m) 1.77 x 10-6 m2 R = 7.9 x 10-3Ω

  2. Circuits • A closed path where current flows (charges flow) • Has a source of potential difference (cell or battery) • Has a resistance (ex: lightbulb) • Has wires

  3. Resistor: hinders flow of charges Lamp Wire Provides complete path Switch Dry cell Opens and closes circuit path Source of potential difference – pushes charges, causing them to move

  4. Supplies Voltage Opens and closes a circuit Measures Voltage Measures Current Supplies Resistance

  5. Light Bulb Demo http://www.youtube.com/watch?v=csTlRvqSITI&feature=related If one goes out, they all go out

  6. Series Circuits • Has only 1 path for the current to flow • Draw a series circuit with a 9 V battery and 3 resistors reading 100Ω, 300Ω, and 50Ω.

  7. Equivalent (Total) Resistance Req = R1 + R2 + R3 Req = 100 Ω + 300 Ω + 50 Ω Req = 450 Ω

  8. Draw a new equivalent circuit Req = 450 Ω VT = total voltage of circuit

  9. What is the total current? IT = VT RT IT = 9 V 450Ω IT = 0.02 A

  10. Current is constant for a series circuit Therefore: I1 = 0.02 A I2 = 0.02 A I3 = 0.02 A

  11. When solving vertically, use the rules in the ref. tables for series circuits V=IR 2 100 0.02 When solving horizontally, use Ohm’s Law V=IR 6 300 0.02 V=IR 1 50 0.02 450 0.02 9 Now solve for V1, V2, and V3 Check the voltage: VT = V1 + V2 + V3

  12. Draw a 120 V battery in series with 3 resistors. R1 = 2 Ω R2 =4Ω, R3 = 8Ω. 2 Ω 4 Ω 120 V 8 Ω

  13. Find Req, then draw equivalent circuit with 1 resistor and an ammeter. Place directly into circuit. 120 V Req = 14 Ω Current is constant in series so it doesn’t matter where the ammeter is placed Note: Voltmeters get placed around what they are measuring

  14. Solve for all currents and voltages V1 = I1R1 V1 = (8.57)(2) 17.14 IT = I1 = I2 = I3 8.57 2 V2 = I2R2 V1 = (8.57)(4) 34.28 IT = I1 = I2 = I3 8.57 4 V3 = I3R3 V1 = (8.57)(8) 68.56 IT = I1 = I2 = I3 8.57 8 R1+R2+R3 2+4+8 14 IT =VT / RT IT =120/14 8.57 120 V1+V2+V3 17.14+34.28+68.56 Remember to check the voltages!

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