2 Warmup: On the circuit drawing below label conventional current flow and electron flow. Why is conventional current flow technically the “wrong” way?
3 I = q q = It t q = (3A)(12s) tI = qt t q = 36 C q = It How many electrons flow through a battery that delivers a current of 3 A for 12 s? (HINT: 1 C = 6.25 x 1018 e)I = qtq = Itq = (3A)(12s)tI = qttq = 36 Cq = It#e = (36C)(6.25 x 1018e)#e = 2.25 x 1020
4 Resistance(V) Electric potential difference (voltage- like a battery) isnecessary for current to flow.(Electric Pressure).(I) Current is the flow of electrons through a circuit.(R) Resistance discourages or controls the flow.It is caused by “things” that get in the way of a direct path.The resistance of a conductor (like a wire) is the ratio of thepotential difference applied to the circuit and the current thatflows through it…
5 This is Ohm’s Law…R = VIR = VoltsAmpsR = ohm (Ω)
6 Resistors are “things” in a circuit that limit current flow Resistors are “things” in a circuit that limit current flow. They can be controlled resistors or electrical devices like light bulbs or lamps.
7 Circuit AnalogyR = VIThe pipe is the counterpart of the wire in the electric circuit.The pump is the mechanical counterpart of the battery.The pressure generated by the pump, that drives the water through the pipe, is like the voltage generated by the battery to drive the electrons through the circuit.The seashells plug up the pipe and constrict the flow of the water creating a pressure difference from one end to the other. In a similar manner the resistance in the electric circuit resists the flow of electricity and creates a voltage drop from one end to the other. Energy is lost across the resistor and shows up as heat.
8 Try this…A student measures a current of .10 A flowing through a lightbulb connected by short wires to a 12 V battery.What is the resistance of the light bulb?R = VIR = 12 V.10 AR = 120 Ω
9 Try this… A lamp with a resistance of 20 Ω is in a circuit that has a current of .05 A flowing through it. What is the potentialdifference across the lamp?R = VIV = I RV = (.05 A)(20 Ω)I R = V IIV = 1 VV = I R
10 Recall slope…What are the slopes for the following graphs? spring constantvelocitydFv = dtk = FxtxaccelerationmassvFa = vtm = Fata
11 Guess what the slope is for this graph: VResistance (R)R = VII
12 Wires made of a certain types of metals are used in circuits because they are good conductors and allow the electrons within them to move relatively freely.Although wires allow current to flow through a circuitthey also control the current or have some resistance.
13 Factors that affect the resistance of a wire: LENGTH: Increasing the length (L) of the wire willincrease the resistance of the wire.This is because the current (electrons) will now have furtherto travel and will encounter and collide with an increasingnumber of atoms.
14 Factors that affect the resistance of a wire: 2. AREA: Increasing the cross-sectional area (A) of a wire willdecrease the wires resistance.A = πr2This occurs because in making the wire thickerthere is now more spaces between atoms throughwhich the electrons can travel and thus flow easier.The wire isn’t resisting the flow as much.
15 Factors that affect the resistance of a wire: RESISTIVITY (ρ) – this is a characteristic of a materialthat depends on its electronic structure and temperature. If awire is made of a material that has a high resistivity then it willhave a high resistance.
16 Combining all these factors gives us the following equation for the resistance of a wire: R = ρLAShort/thick/cold wires = low resistance (easy for electrons to flow)Long/thin/hot wires = high resistance (hard for electrons to flow)
17 Try this… R = ρL A = (1.72 x 10-8 Ω•m)(4m) π(.001m)2 = .0219 Ω Determine the resistance of a 4.00 m length of copper wirehaving a diameter of 2 mm. Assume the temperature of 20°C.4 md= 2 mmρ copper = 1.72 x 10-8 Ω•mR = ρLA= (1.72 x 10-8 Ω•m)(4m)π(.001m)2= Ω
18 Determine the length of a copper wire that has a resistance of Determine the length of a copper wire that has a resistance of .172 Ω and cross-sectional area of 1 x 10-4 m2. The resistivity of copper is x 10-8 Ω•m.L = ARρAR = ρLAAL = (.001m2)(.172Ω)(1.72 x 10-8 Ω•m)AR = ρLρ ρL = 10,000 mL = ARρ
19 Wire A Which one of the five wires has the largest resistance? MaterialLengthDiameterAiron2 m6.4 x 10-4 mBcopperC1.2 x 10-3 mD1 mEIronρ iron = 9.7 x 10-8 Ω•mρ copper = 1.7 x 10-8 Ω•mWhich one of the five wires has the largest resistance?High resistance = long/thin/hot wiresWire A
20 Wire D Of the five wires, which one has the smallest resistance? MaterialLengthDiameterAiron2 m6.4 x 10-4 mBcopperC1.2 x 10-3 mD1 mEIronρ iron = 9.7 x 10-8 Ω•mρ copper = 1.7 x 10-8 Ω•mOf the five wires, which one has the smallest resistance?Small resistance = short/thick/coldWire D
21 WireMaterialLengthDiameterAiron2 m6.4 x 10-4 mBcopperC1.2 x 10-3 mD1 mEIronρ iron = 9.7 x 10-8 Ω•mρ copper = 1.7 x 10-8 Ω•mWhich of the wires carries the smallest current when they are connected to identical batteries?How does current (I) relate to resistance (R)?R = VIWhen current is low, resistance is high.Wire AHigh resistance = long/thin/hot wires
22 Resistance Variables Lab Coil #MetalLength(meters)CrossSectionalArea_____GaugeVVolts(V)ICurrent(A)RResistance(Ω)1Copper10 m.325 mm2222.081 mm228320 m45Copper-Nickelρ copper = 1.7 x 10-8 Ω•mρ cop-nick = 4.9 x 10-8 Ω•mFirst use R = ρL/A to predict the resistance of each circuit.