2/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 1/3Master ItConsider the circuit shown in the figure below. (R = 27.0 Ω.)(a) Find the current in the 27.0 Ω resistor.(b) Find the potential difference between points a and b.Part 1 of 7 ConceptualizeThe current from b to a around the outer loop through R and the 5Ω resistor is on the order of fractions ofamps. Point b will be at higher potential than point a by several volts.Part 2 of 7 CategorizeWe must analyze the entire circuit in order to find the current in any one resistor. This circuit has a singlepower supply so we can reduce the circuit into equivalent circuits with equivalent resistances for series andparallel combinations.Part 3 of 7 AnalyzeIf we turn the diagram of the original figure on its side and shorten some of the wires, we see that it is thesame as diagram (i) in the figure below. Resistances denoted by and in the diagrams belowrefer to the equivalent resistances in these circuits. Since diagram (i) is merely a rotation of the originalfigure, the unknown resistance here is denoted by R as in the original figure. The resistor R and the 5Ωresistor are in series, so the first reduction into an equivalent circuit is shown in diagram (ii). For theequivalent resistance in this circuit, we haveIn diagram (ii), since the 10.0Ω, the 5.00Ω, and resistor are in parallel, we compute the equivalentresistance asRii, Riii, RivRii = 5.00 Ω + R = 32 Ω.Rii2/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 2/3Part 4 of 7 AnalyzeThis is shown in diagram (iii) which reduces to the circuit shown in diagram (iv). The two resistances indiagram (iii) are in series, so we havePart 5 of 7 AnalyzeNow we work backwards through the equivalent circuit diagrams applying and ΔV = IR alternatelyto each resistor, real or equivalent. The resistor is connected across 25.0 V, so the current through thevoltage source in every diagram is given byPart 6 of 7 AnalyzeIn diagram (iii), the current above goes through the equivalent resistor to give a voltage drop across thisresistor ofPart 7 of 7 AnalyzeIn diagram (ii), we see that this voltage drop is and is the same across the 10.0Ω resistor and the5.00Ω resistor.(a) Since the current through the resistor R is also the current through the 5.00Ω resistor, the combinationbeing connected between points a and b, this current is given by(b) We already found the potential difference above between points a and b to beRiii = = 3.02 Ω.11/10.0 Ω + 1/5.00 Ω + 1/ 32 ΩRiv = Riii + 10.0 Ω = 13 Ω.I = ΔV/RRivI = = = 1.92 A.ΔVRiv 25 V 13 ΩRiiiΔV = IRiii = 1.92 A 3.02 Ω = 5.8 V.ΔVabI = = = = 0.181 A.ΔVabRabΔVab(R + 5.00 Ω) 5.8 V 32 Ω2/23/2014 Master Ithttp://www.webassign.net/v4cgi/questions/tutorial_popup.tpl 3/3FinalizeWe could not use in the original circuit because there was no resistor for which we knew either ofthese quantities. It was necessary to draw all of the equivalent circuits in diagrams (i) through (iv) and workbackward to find the current across each resistor. Both answers are smaller than our estimates. Most of thecurrent goes through the 5.00 Ω resistor in the middle of the original figure.ΔVab = 5.8
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