Factors affecting resistance

1. Factors Affecting Resistance Reported By: Ralph Lery Guerrero Kevin Roxas Marco Lauro Delos Santos

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. 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. Length R 1 /R 2 =l 1 /l 2 Where R1= resistance of the first conductor(in ohms) R2= resistance of the second conductor(in ohms) l1=length of the first conductor (in cm) l2=length of the second conductor (in cm)

5. Length Resistivity- is the resistance per unit length of a specific substance to electric force.

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. 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. 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. Diameter (Cross-sectional area) R 1 /R 2 =d 2 2 /d 1 2 Some 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. 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: R 2

11. Diameter (Cross-sectional area) Given: R 1 =0.4Ω d 1 =0.8mm d 2 =0.4mm Solution: R 1 /R 2 =d 2 2 /d 1 2 0.4Ω/R 2 =(0.4mm) 2 /(0.8mm) 2 R 2 =(0.4Ω)(0.64mm) 2 /0.16mm 2 R 2 =1.60Ω

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. Temperature Resistance increases with the temperature.

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. Resistors- conductors whose resistance is constant when the temperature is constant. Kind of Material (Resistivity)

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 Alloys Constantan (alloy of Cu and Ni) 49 x 10 -6   Manganin (alloy of Cu, Mn and Ni) 44 x 10 -6   Nichrome (alloy of Ni, Cr, Mn and Fe) 100 x 10 -6 Insulators Glass 10 10  - 10 14   Hard rubber 10 13  - 10 16   Ebonite 10 15  - 10 17   Diamond 10 12  - 10 13   Paper (dry) 10 12  

17. Kind of Material (Resistivity) Problem Given: Find: L=15 m R Diameter=0.085 cm = 1.6x10 -8 Ωm T=20°C Solutions: Conversion: d= 0.085cm/100 cm x 1m d= (8.5x10 -4 m)

18. a.) r=1/2d r=8.5x10 -4 m/2 r= 4.25x10 -4 m b.) A=πr 2 A=(3.14)(4.25x10 -4 m) 2 A=5.67x10 -7 Kind of Material (Resistivity)

19. Kind of Material (Resistivity) c.) R = L A R =(1.6x10 -8 Ωm)(15m) 5.67x10 -7 m 2 R= 0.4Ω

20. Thank you! =)) Signing Off … By: RLUG