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## Resistance

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**Resistance**and resistivity**Current**• Current is sort of a vector • Direction is constrained by conductor • Restricted to forward or backward (+ or –)**Resistance**• Current does not flow unhindered • Electrical resistance is analogous to friction or drag • Expressed as potential needed to maintain a current**Ohm’s Law**DV I = R I = current DV = voltage = electric potential drop R = resistance Unit of resistance : V / A = ohm (W)**Voltage Causes Current**Potential drop is the cause. Current is the effect. Resistance reduces the effect of potential.**Does it Work?**• Approximation of varying utility:R is independent of DV and I • When true, the material is ohmic**Poll Question**If you want to increase the current through a resistor, you need to • Increase the resistance or voltage. • Decrease the resistance or voltage. • Increase the resistance or decrease the voltage. • Decrease the resistance or increase the voltage.**Ohm’s Law Rearranged**If you know two, you can find the third. DV DV I = R = DV= IR R I I = current DV = potential R = resistance**Example**A 1.5-V battery powers a light bulb with a resistance of 9 W. What is the current through the bulb? Ohm’s Law I = V / R V = 1.5 V; R = 9 W I = (1.5 V ) / (9 V/A) = 1/6 A**Resistivity**For current through a cylinder: L A • Longer L greaterR. • Greater A smallerR. • More resistive material bigger R.**Resistivity**• R = rL/A • ris Resistivity • Unit: ohm·meter = Wm • More or less constant depending on material, conditions**Resistivity**• Intensive quantity • Does not depend on the amount of material, only its conditions • Predictive value when mostly constant (ohmic)**Example**The resistivity of copper is 1.710–8Wm. What is the resistance of a 100-km length of copper wire that is 1/4” in diameter?**Classes of Conductors**• How resistivity changes with temperature a = temperature coefficient of resistivity**Classes of Conductors**• How resistivity changes with temperature**Power**dissipated by a resistor**Electric Power**Potential is energy per charge:V = DE/ q Current is charge per time:I= q /Dt So, (potential times current) = (energy per time) = power Power = VI**Group Work**Power P = VI and V = IR. Using these, show that: • P = I2R • P = V2/R