Factors affecting resistance

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Factors affecting resistance

  1. 1. Factors Affecting ResistanceReported By: Ralph Lery Guerrero Kevin Roxas Marco Lauro Delos Santos
  2. 2. Resistance• Is defined as an obstacle to the flow of electric current.• Is the opposition offered by any object to the passage of an electric current through it.
  3. 3. Length• Resistance of a conductor is directly proportional to the length of the wire i.e., longer the wire greater will be the resistance and shorter the wire smaller will be the resistance. If L represents the length of the uniform wire, then RL
  4. 4. Length R1/R2=l1/l2Where R1= resistance of the first conductor(in ohms) R2= resistance of the second conductor(inohms) l1=length of the first conductor (in cm) l2=length of the second conductor (in cm)
  5. 5. Length• Resistivity- is the resistance per unit length of a specific substance to electric force.
  6. 6. Length• The volume control of your radio is a variable resistor. As you turn the volume control knob, the effective the length of the resistance changes.• Resistance and the amount of current passing through the circuit change, making the sound from the speaker soft and loud.
  7. 7. Diameter (Cross-sectional area)• Resistance of a conductor is inversely proportional to the area of the cross- section of the uniform wire. That means, thinner the wire, greater the resistance and thicker the wire, lower the resistance. If A is the area of cross- section of the uniform wire, then,
  8. 8. Diameter (Cross-sectional area) If the cross-sectional area is doubled, twice as many electrons became available to flow so that the current is doubled.
  9. 9. Diameter (Cross-sectional area) R1/R2=d22/d12Some connecting wires are made of stranded thin wires.Stranding wires has the same effect as increasing cross-sectional area.The wire’s gauge number specifies its size. The smaller gauge number, the thicker the wire. Hence, the lesser the resistance of the conductor.
  10. 10. Diameter (Cross-sectional area)Problem An iron wire conductor with a diameter of 0.8 mm has a resistance of 0.4 ohm Ω. Calculate the resistance of a n iron wire 0.4 mm in diameter.Find: R2
  11. 11. Diameter (Cross-sectional area)Given: R1=0.4Ω d1=0.8mm d2=0.4mmSolution:R1/R2=d22/d120.4Ω/R2=(0.4mm)2/(0.8mm)2R2=(0.4Ω)(0.64mm)2/0.16mm2R2=1.60Ω
  12. 12. Temperature• The resistance of a metallic conductor increases as the temperature increases e.g. copper• The resistance of a semiconductor/insulator decreases as the temperature increases
  13. 13. TemperatureResistance increases with the temperature.
  14. 14. Kind of Material (Resistivity)• The material also affects the resistance of a conductor by a fixed amount for different materials. This is known as resistivity (). R = L  = constant of proportionality A
  15. 15. Kind of Material (Resistivity)Resistors- conductors whose resistance is constant when the temperature is constant.
  16. 16. Kind of Material (Resistivity) Material Resistivity (W m)Conductors Silver 1.60 x 10-8 Copper 1.62 x 10-8 Aluminium 2.63 x 10-8 Tungsten 5.20 x 10-8 Nickel 6.84 x 10-8 Iron 10.0 x 10-8 Chromium 12.9 x 10-8 Mercury 94.0 x 10-8 Manganese 1.84 x 10-6 Constantan (alloy of CuAlloys 49 x 10-6 and Ni) Manganin (alloy of Cu, Mn 44 x 10-6 and Ni) Nichrome (alloy of Ni, Cr, 100 x 10-6 Mn and Fe)Insulators Glass 1010 - 1014 Hard rubber 1013 - 1016 Ebonite 1015 - 1017 Diamond 1012 - 1013 Paper (dry) 1012
  17. 17. Kind of Material (Resistivity)ProblemGiven: Find:L=15 m RDiameter=0.085 cm= 1.6x10-8 ΩmT=20°CSolutions:Conversion: d= 0.085cm/100 cm x 1m d= (8.5x10-4m)
  18. 18. Kind of Material (Resistivity)a.) r=1/2d r=8.5x10-4 m/2 r= 4.25x10-4 mb.) A=πr2 A=(3.14)(4.25x10-4 m)2 A=5.67x10-7
  19. 19. Kind of Material (Resistivity)c.) R = L A R=(1.6x10-8Ωm)(15m) 5.67x10-7m2 R=0.4Ω
  20. 20. Signing Off … By: RLUGTHANK YOU! =))

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