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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 vso 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 75W 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.
 Spring '08
 LUO
 Physics

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