Lecture 15: MON 16 FEBLecture 15: MON 16 FEBDC CircuitsDC CircuitsCh27.1Ch27.1––44Physics 2102Jonathan DowlingbabaThe battery operates as a “pump” thatmoves positive charges from lower to higherelectric potential. A battery is an exampleof an “electromotive force” (EMF) device.These come in various kinds, and all transform one source of energy into electricalenergy. A battery uses chemical energy, a generator mechanical energy, a solarcell energy from light, etc.The difference in potentialenergy that the deviceestablishes is called the EMFand denoted by E.EMF Devices and Single-Loop CircuitsEMF Devices and Single-Loop CircuitsE = iRa b c d=aVaEiRbadc- +iiGiven the EMF devices and resistors in acircuit, we want to calculate thecirculating currents. Circuit solvingconsists in “taking a walk” along thewires. As one “walks” through thecircuit (in any direction) one needs tofollow two rules:When walking through an EMF, add +E if you flow with the current or–E against. How to remember: current “gains” potential in a battery.When walking through a resistor, add -iR, if flowing with the current or +iRagainst. How to remember: resistors are passive, current flows “potential down”.Example:Walking clockwise from a: +E–iR=0.Walking counter-clockwise from a: -E +iR=0.Circuit ProblemsCircuit ProblemsIf one connects resistors of lower and lower value of R to get higherand higher currents, eventually a real battery fails to establish thepotential difference E, and settles for a lower value.One can represent a “real EMF device” as an ideal one attachedto a resistor, called “internal resistance” of the EMF device:The true EMF is a function of current: the morecurrent we want, the smaller Etrue we get.Ideal Ideal vsvs. Real Batteries. Real Batteries ! Etrue= E " ir ! E " ir " iR = 0 # i = E/ r + R( )Resistances in Series:Resistances in Series: i is Constanti is ConstantTwo resistors are “in series” if they are connected such that thesame current i flows in both.The “equivalent resistance” is a single imaginary resistor that canreplace the resistances in series.“Walking the loop” results in :In the circuit with theequivalent resistance,Thus, !==njjeqRR1 ! E " iR1" iR2" iR3= 0 # i = E/ R1+ R2+ R3( ) ! E " iReq" = 0 # i = E/ReqResistors in Parallel: V is ConstantResistors in Parallel: V is ConstantTwo resistors are “in parallel” ifthey are connected such that thereis the same potential V dropthrough both. The “equivalentresistance” is a single imaginaryresistor that can replace theresistances in parallel.“Walking the loops” results in:The total current delivered by thebattery is:In the circuit with the equivalentresistor,!==njjeqRR111 ! E " i1R1= 0, E " i2R2= 0, E " i3R3= 0. ! i = i1+ i2+ i3= E / R1+ E / R2+ E / R3= E 1/ R1+ 1/ R2+ 1/ R3( ) ! i = E / Req! V = iR! Q = CVResistorsCapacitorsSeries: I!=!dQ/dt SameSeries: Q SameParallel: V SameParallel: V Same! Rser= R1+ R2+ R3+ ...! 1/Rpar= 1/R1+ 1/R2+ 1/R3+ ...! 1/Cser= 1/C1+ 1/C2+ 1/C3+ ...! Cpar= C1+ C2+ C3+ ...Resistors in Series and ParallelResistors in Series and ParallelAn electrical cable consists of 100 strands of fine wire, each having r=2Ω resistance. The samepotential difference is applied between the ends of all the strands and results in a totalcurrent of I=5 A.(a)What is the current in each strand?Ans: ip=0.05 A (i=I/100)(b)What is the applied potential difference?Ans: vp=0.1 V (vp=V=isr=constant)(c)What is the resistance of the cable?Ans: Rp=r=0.02Ω (1/Rp=1/r+1/r+…=100/r => R=r/100)Assume now that the same 2 Ω strands in the cable are tied in series, one after the other, andthe 100 times longer cable connected to the same V=0.1 Volts potential difference as before.(d)What is the potential difference through each strand?Ans: vs=0.001 V (vs=V/100)(e)What is the current in each strand?Ans: is=0.0005 A (is=vs/r=constant)(f)What is the resistance of the cable?Ans: 200 Ω (Rs=r+r+r+…=100r)(g)Which cable gets hotter, the one with strands in parallel or the one with strands in series?Ans: Each strand in parallel dissipates Pp=ivp=5mW (and the cable dissipates 100•Pp=500mW);Each strand in series dissipates Ps=is•vs=50 µW (and the cable dissipates 5mW)…ParallelSeriesAVRVi 5.1812=!==ExampleExampleBottom loop: (all else is irrelevant)V same in parallel!8W12VWhich resistor (3 or 5) gets hotter? P=i2RExampleExamplea) Which circuit hasthe largestequivalentresistance?b) Assuming that allresistors are thesame, which onedissipates morepower?c) Which resistor hasthe smallestpotential differenceacross it?ExampleExampleFind the equivalent resistance between points(a) F and H and(b) F and G.(Hint: For each pair of points, imagine that abattery is connected across the pair.)Monster MazesMonster MazesIf all resistors havea resistance of 4Ω,and all batteries areideal and have anemf of 4V, what isthe current throughR?If all capacitors havea capacitance of 6µF,and all batteries areideal and have an emfof 10V, what is thecharge on capacitor
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