Ohm’s LawThe relationship between the potentialdifference in a circuit and the resultingcurrent.IRVIRV=∆=∆or
20.1 Electromotive Force and CurrentThe electric current is the amount of charge per unit time that passesthrough a surface that is perpendicular to the motion of the charges.http://phet.colorado.edu/en/simulation/signal-circuit
20.1 Electromotive Force and CurrentThe electric current is the amount of charge per unit time that passesthrough a surface that is perpendicular to the motion of the charges. Byconvention, we use the flow of positive charge or “electron holes”.tqI∆∆=One coulomb per second equals one ampere (A).
20.2 Ohm’s LawTo the extent that a wire or anelectricaldevice offers resistance toelectrical flow,it is called a resistor.
Kirchoff’s Junction LawThe sum of thecurrents enteringa junction areequal to the sumof the currentsleaving.
What current flows through the 4thwire and whichway does it go?A.1 A into the junctionB.1 A out of the junctionC.9 A into the junctionD.9 A out of the junction
In which direction does the potentialdecrease?A.ClockwiseB.CounterclockwiseC.Potential is the same everywhere in thecircuit.
Ohm’s Law in the CellThe resistance between the walls of a biological cell is 1.00x 1010Ω.(a)What is the current when the potential differencebetween the walls is 95 mV?(b)If the current is composed of Na+ions (q = +e), howmuch charge moved between the cell walls in 0.2 s?
Ohm’s Law in the CellSuppose that the resistance between the walls of abiological cell is 1.00 x 1010Ω.(a)What is the current when the potential differencebetween the walls is 95 mV? 9.5 x 10-12A(b)If the current is composed of Na+ions (q = +e), howmuch charge moved between the cell walls in 0.2 s? 1.9x 10-12C
Power in a circuitThrough which resistor is the most powerdissipated?A. a B. b C. b,d D. d
Power rating of a household applicance• Commercial and residential electrical systems are set upso that each individual appliance operates at a potentialdifference of 120 V.• Power Rating or Wattage is the power that the appliancewill dissipate at a potential difference of 120 V (e.g. 100W bulb, 1000 W space heater).• Power consumption will differ if operated at any othervoltage.• Energy is often expressed as kilowatt-hours, for meteringpurposes.
kW-hHow much energy in a kilowatt-hour?A)1000 JoulesB)0.28 JoulesC)3.6 million JoulesD)60,000 Joules
Resistors in Series• Series wiring means that the resistors are connected so that there isthe same current through each resistor.• The potential difference through each resistor is :|∆VR |= IR• The equivalent resistor (Rs ) is made by adding up all resistors inseries in the circuit.∆Vbat = |∆Vcircuit |= IRs
20.6 Series Wiring•By understanding that thepotential difference across anycircuit resistor will be a voltageDROP, while the potentialdifference back through the battery(charge escalator) will be a GAIN,one can replace the ΔV with “V” ,and dispense with the absolutevalue signs.V1 (voltage drop across theresistor1) = IR1V2 (voltage drop across theresistor2) = IR2Vbat = Vcir = IReq
20.6 Series WiringResistors in a Series CircuitA 6.00 Ω resistor and a 3.00 Ω resistor are connected in series witha 12.0 V battery. Assuming the battery contributes no resistance tothe circuit, find (a) the current, (b) the power dissipated in each resistor,and (c) the total power delivered to the resistors by the battery.
Series ResistorsWhat is the value of R (3rdresistor in circuit)?a. 50 Ω b. 25Ω c. 10 Ω d. 0
20.7 Parallel WiringParallel wiring means that the devices areconnected in such a way that the samevoltage is applied across each device.When two resistors are connected inparallel, each receives current from thebattery as if the other was not present.Therefore the two resistors connected inparallel draw more current than does eitherresistor alone.
20.7 Parallel Wiring
20.7 Parallel WiringThe two parallel pipe sections are equivalent to a single pipe of thesame length and same total cross sectional area.
20.7 Parallel Wiringparallel resistors+++=3211111RRRRPnote that for 2 parallel resistors thisequation is equal to :Rp = (R1R2 ) / (R1 + R2)Note that the potential difference ofthe battery is the same as thepotential difference across eachresistor in parallel.
20.7 Parallel WiringExample 10 Main and Remote Stereo SpeakersMost receivers allow the user to connect to “remote” speakers in additionto the main speakers. At the instant represented in the picture, the voltageacross the speakers is 6.00 V. Determine (a) the equivalent resistanceof the two speakers, (b) the total current supplied by the receiver, (c) thecurrent in each speaker, and (d) the power dissipated in each speaker.
Circuits Wired Partially in Series and Partially in ParallelExample 12Find a) the totalcurrent suppliedbythe the batteryand b) thevoltage dropbetween points Aand B.
Circuits Wired Partially in Series and Partially in ParallelExample 12Find a) thetotal currentsupplied bythe thebattery andb) thevoltagedropbetweenpoints Aand B.
Circuits Wired Partially in Series and Partially in ParallelExample 12a) Itot = V/Rp = 24V/240Ω = .10Ab) now go back to the 1st circuit in part c tocalculate the voltage drop across A-B:VAB = IRAB = (.10A)(130 Ω) = 13V
Your Understanding• What is the ratio ofthe power suppliedby the battery inparallel circuit A tothe power suppliedby the battery inseries circuit B?a.¼ c. 2 e. 1b. 4 d. 2A.B.
Analyze the identical bulbsWhich bulbs will lighta.All bulbsb.A, Bc.A onlyd.nonehttp://bcs.wiley.com/he-bcs/Books?action=resource
What happens to A?With switch closed, Ais______________it was whenswitch was open.a.Brighter thanb.Less bright thanc.Equal to what http://bcs.wiley.com/he-bcs/Books?action=resoclosed
Complex CircuitThe current throughthe 8 Ω resistor is .590A. Determine thefollowing:a.Current in the 20 Ωresistorb.Current in the 9 Ωresistorc.Battery voltage
Complex CircuitThe current throughthe 8 Ω resistor is .590Determine thefollowing:a.Current in the 20 Ωresistor: .885 Ab.Current in the 9 Ωresistor: 2.49 Ac.Battery voltage: 22.4V
20.11 The Measurement of Current and VoltageAn ammeter must be inserted intoa circuit so that the current passesdirectly through it.The resistance of the ammeterchanges the current through thecircuit.The ideal ammeter has a very lowresistance
20.11 The Measurement of Current and VoltageTo measure the voltage between twopoints, in a circuit, a voltmeter is connectedin parallel, between the points.A voltmeter takes some current away fromthe circuit it measures.The ideal voltmeter has a very largeresistance so it diverts a negligible current.
• A battery consists ofchemicals, called electrolytes,sandwiched in between 2electrodes, or terminals, madeof different metals.• Chemical reactions do positivework and separate charge.The electric field does thesame amount of negativework, which translates as apotential difference betweenpositive and negative terminals• A physical separator keeps thecharge from going backthrough the battery.The Battery
ε and ∆V• For an ideal battery potentialdifference between the positiveand negative terminals, ∆V,equals the chemical work doneper unit charge.∆V = Wch /q = ε• ε is the emf of the battery• Due to the internal resistance ofa real battery, ∆V is oftenslightly less than the emf.• A capacitor can store charge, buthas no way to do the work toseparate the charge. It has apotential difference, but no ε.
How do we know there is acurrent?
20.1 Electromotive Force and CurrentIf the charges move around the circuit in the same direction at all times,the current is said to be direct current (dc).If the charges move first one way and then the opposite way, the current issaid to be alternating current (ac).
EOC #3A fax machine uses 0.110 A of current in its normal modeof operation, but only 0.067 A in the standby mode. Themachine uses a potential difference of 120 V.(a)In one minute (60 s!), how much more charge passesthrough the machine in the normal mode than in thestandby mode?(b)How much more energy is used?
EOC #3A fax machine uses 0.110 A of current in its normal modeof operation, but only 0.067 A in the standby mode. Themachine uses a potential difference of 120 V.(a)In one minute (60 s!), how much more charge passesthrough the machine in the normal mode than in thestandby mode? .043 C/s x 60 s = 2.6 C(b)How much more energy is used? 2.58 C x 120 J/C = 310J
20.1 Electromotive Force and CurrentConventional current is the hypothetical flow of positive charges (electronholes) that would have the same effect in the circuit as the movement ofnegative charges that actually does occur.
20.2 Ohm’s LawThe resistance (R) is defined as theratio of the voltage V applied acrossa piece of material to the current I throughthe material.
20.2 Ohm’s LawTo the extent that a wire or an electricaldevice offers resistance to electrical flow,it is called a resistor.
20.2 Ohm’s LawExample 2 A FlashlightThe filament in a light bulb is a resistor in the formof a thin piece of wire. The wire becomes hot enoughto emit light because of the current in it. The flashlightuses two 1.5-V batteries to provide a current of0.40 A in the filament. Determine the resistance ofthe glowing filament.Ω=== 5.7A0.40V0.3IVR