solution_Ch18

solution_Ch18 - (a) When the three resistors are in series,...

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Unformatted text preview: (a) When the three resistors are in series, the equivalent resistance of the circuit is Rm :R1+R3+R3 = 3(9.0 Q): The terminal potential difference of the battery is applied across the series combination of the three 9.0 Q resistors, so the current supplied by the battery and the current through each resistor in the series combination is av 12v :—:—:-0.44A R 279 as 1 If the three 9.0 Q are now connected in parallel with each other, the equivalent resistance is l l l l 3 —:7—+—7 +—, 2—7 7 R 9.0 9 9.0 9 9.0 $2 9.0 $2 en 01‘ When this parallel combination is connected to the battery. the potential difference across each resistor in the combination is AV = l2 V, so the current through each of the resistors is 1.3A The parallel combination of the 6.0 Q and 12 Q resistors has an equivalent resistance of l 1 2+1 _12r2_ , + , = , 6.082 1282 1282 3 4.0 Q or ,21 pl Similarly, the equivalent resistance ol‘ the 4.0 $2 and 8.0 $2 parallel combination is l l l 2+1 + : 4.0 Q 8.0 Q 8.0 Q R p: The total resistance of the series combination between points a and b is then RM :R‘,,I+S,OQ+Rp2 :4.OQ+5,09+8,—%0 Q: If AV“, : 35 V. the total current from a to b is [(1,], : AVHh/Rm, : 35 V/(35 (2/3) : 3.0 A and the potential differences across the two parallel combinations are AVN : [MR/,1 : (3.0 A)(4.0 Q) : 12 V and AV”: : [1,,th so the individual currents through the various resistors are: 1.0A; 1.0A; Il’leVul/IZQ: IfiZAVN/ODQ: Ix : AVG/8,0 Q : and I4 : AVG/4,0 Q : Observe that the center branch of this circuit. that is the branch containing points (I and b, is not a continuous conducting path. so no current can flow in this branch. The only current in the circuit flows counterclockwise around the perimeter of this circuit Going counterclockwise around the this outer loop and applying Kirchhot‘f‘s loop rule gives 78.0 V410 (2)1780 Q)l+l2 who o)17(5.09)120 712 v-so v 7 209 Now, we start at point I) and go around the upper panel of the circuit to point a. keeping track of changes in potential as they occur. This gives or I 0.20 A AV : v —V,, : —4.0 v+(6.oo)(0)—(3.0 o)(0.20 A)+12 V—(lO r2)(0.20 A) :+5.4 v all u Since AV”) > 0, point a is 5.4 V higher in potential than point b , Following the path of I, from (i to 17. and recording changes in potential gives V 11,: +24 vit'oo Q)(3.0 A) :+o.0 v 1:7 Now, following the path of I 3 from (I to b, and recording changes in potential gives i141, = 7(340 (2)11 =+o.0 v. or 12 z Thus, I2 is directed from and has magnitude of 2.0 A. Applying Kirchhoff’s junction rule at point (1 gives 13 :11+12 : 3.0 A+(—2_0 A): . . . . I: 0.600A The total power input to the ClI‘Cult is (P : (isI +531 : (1.50 V+1.50 V)(0.000 A): 1.80 W iupm (p lax. : 13(11 +13): (0000 A): (0.255 Q+0.153 Q) : 0.147 W Thus, the fraction of the power input that is dissipated internally is 1 / ’ ,3“ = 0147 W = 0.0816 0r8.16% <P 1.80 W mpm [8.32 (:1) 1b) = C8:(:20,0><10*” F)(0,00 V’):1.80><10“ C: 18011C ., / 1 1c) - /“‘)=QWL1*-]= 6/ 13.33 = C8 = (5.0 x10” F)(30 V) = 1.5 x10” C. and 1' = RC = (1.0 x10° £2)(5.0 x10” F) = 5.0 s Thus,aL/=lOs=21' elm—r): _ 8 . . (a) 1W = so the resrstance is 48.0 V :fizquto“ Q 0.300xl0' A R:— 1 max The time constant is T : R C. so the capacitance is found to be T 0960 s - C:—:—:l00><l0 " F: l0.0 F R away 9 (b) QMx : C8 : (10.0 ‘uF)(48.0 Vi) : 480 ,LLC. so the charge stored in the capacitor at t : 1.92 s is 1,92 ,3 The current drawn by a single 75-W bulb connecth to a lZO—V source is I1 : (P/AV : 75 W/120 V. Thus. the number of such bulbs that can be connected in parallel with this source before the total current drawn will equal 30.0 A is 30.0 A ’120 V , r : : 30.0A : 42; n I l )[ 75 W) l The resistive network between a an b reduces. in the stages shown below, to an equivalent resistance of ch : . 2.40 5.19 ...
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This note was uploaded on 03/30/2009 for the course PHY 102 taught by Professor Luo during the Spring '08 term at SUNY Buffalo.

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solution_Ch18 - (a) When the three resistors are in series,...

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