ࡱ> ,.+S7 bjbjUU "D7|7|bl>>>>...B^^^^4LB\LLL[[[[[[[$] _[.LJLLL[N>>93^\NNNL4>H.[NL[NZ NY.Y m"gB^LYYt\0\Yd`iMd`YNBB>>>>Physics Challenge Question 21: Solutions Part 1 As can be seen on the diagram, the 3 W and the 2 W resistors are in parallel. These two resistors are in series with the 5 W resistor. (An electron flowing through the circuit can go through either the 3 W or the 2 W resistor, but after that it has to go through the 5 W resistor.) Part 2 Since the 3 W and the 2 W resistors are in parallel, their effective resistance is  EMBED Equation.3  Cross-multiplying gives  EMBED Equation.3  This  effective resistor, as mentioned in part 1, is in series with the 5 W resistor, so the total resistance of the circuit is  EMBED Equation.3  Part 3 Let us first find the current flowing in the circuit. We can use Ohms law:  EMBED Equation.3  This is the current flowing throughout the circuit. Therefore, 0.97 A is also the current flowing through the 5 W resistor. However, it splits up into I1 and I2 when reaching the two resistors in parallel, as shown on the diagram to the right. Let s summarize what we know so far: VIR3 W resistor3.00 W2 W resistor2.00 W5 W resistor0.97 A5.00 WTotal6.00 V0.97 A6.20 WLooking at our table, we can use Ohm s law to find the potential difference over the 5 W resistor:  EMBED Equation.3  Since the 5 W resistor is in series with the two others, that means the remaining voltage must be over the 3 W and the 2 W resistors. Notice that since the 3 W and the 2 W resistors are in parallel with each other, they both have the same potential difference.  EMBED Equation.3   EMBED Equation.3  Now that we know the voltage over the two remaining resistors, we can use Ohm s law to find the current through each of them: The 3 W:  EMBED Equation.3  The 2 W:  EMBED Equation.3  Notice that the 3 W resistor gets less current. Since it is harder to go through that one, most of the current goes through the 2 W resistor instead. Notice also that these two currents add up to the total (0.97 A), just like they should; no current is lost in the circuit. We have now completely filled in the table: VIR3 W resistor1.16 V0.39 A3.00 W2 W resistor1.16 V0.58 A2.00 W5 W resistor4.84 V0.97 A5.00 WTotal6.00 V0.97 A6.20 W Part 4 Adding many resistors in parallel makes it easier for the current to flow. If I keep adding resistors in parallel, it eventually becomes  infinitely easy for the current to flow. (It has more ways to go, which lowers the resistance.) This can also be seen from the equation:  EMBED Equation.3  Eventually,  EMBED Equation.3  gets really big, so Rtotal itself becomes essentially zero. All that then remains is the 5 W resistor; the total circuit resistance is 5 W. Each of the 1 W resistors has 0 V potential difference, since their total resistance is 0. Since there are so many, each only gets a tiny fraction of the current. (This will be close to 0 A if I just add enough resistors.) (However, notice that the total current in the circuit increases to 1.2 A by Ohm s law.) Part 5 Adding 1 W resistors in series, the total resistance of the circuit would be:  EMBED Equation.3  The total circuit resistance becomes infinitely big. (The current has to go through each new resistor.) Since the resistance becomes infinitely big, no current can flow, and I = 0 A. The total voltage over the entire circuit is still 6 V, but it is now shared by an infinite number of resistors, so they each have a really small voltage. (This will be close to 0 V if I just add enough resistors.) Here are some diagrams that help explain parts 4 (top) and 5 (bottom). 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"TWT5 ; bdgx{(z}z } y | b333333333333v6 A X _`abbccddeeffgMikkel H. JensenC:\Documents and Settings\Elikel\Desktop\Mikkel\Documents\Work\Boston University\GK12 2008-2009\Physics Challenge Questions\Solutions\Solution-CQ21.docMikkel H. JensengC:\Documents and Settings\Elikel\Application Data\Microsoft\Word\AutoRecovery save of Solution-CQ21.asdMikkel H. JensenC:\Documents and Settings\Elikel\Desktop\Mikkel\Documents\Work\Boston University\GK12 2008-2009\Physics Challenge Questions\Solutions\Solution-CQ21.docMikkel H. JensenC:\Documents and Settings\Elikel\Desktop\Mikkel\Documents\Work\Boston University\GK12 2008-2009\Physics Challenge Questions\Solutions\Solution-CQ21.docMikkel H. JensenC:\Documents and Settings\Elikel\Desktop\Mikkel\Documents\Work\Boston University\GK12 2008-2009\Physics Challenge Questions\Solutions\Solution-CQ21.docMikkel H. JensenC:\Documents and Settings\Elikel\Desktop\Mikkel\Documents\Work\Boston University\GK12 2008-2009\Physics Challenge Questions\Solutions\Solution-CQ21.docMikkel H. JensengC:\Documents and Settings\Elikel\Application Data\Microsoft\Word\AutoRecovery save of Solution-CQ21.asdMikkel H. JensengC:\Documents and Settings\Elikel\Application Data\Microsoft\Word\AutoRecovery save of Solution-CQ21.asdMikkel H. JensengC:\Documents and Settings\Elikel\Application Data\Microsoft\Word\AutoRecovery save of Solution-CQ21.asdMikkel H. JensengC:\Documents and Settings\Elikel\Application Data\Microsoft\Word\AutoRecovery save of Solution-CQ21.asd,-.56CDELMZ[bijpw~9:<>@ANU\cdqx@ hC mmmmmmm | | | q@@@@@ @ @@@@@,@@@@8@@UnknownGz Times New Roman5Symbol3& z ArialCFComic Sans MS"qhff#>  !20dM2QHP(Physics Challenge Question 1: SolutionsEditorMikkel H. Jensen