# hw4 - 19.24. 599 19.24 a) The current in the wire is the...

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Unformatted text preview: 19.24. 599 19.24 a) The current in the wire is the same at all points along its length, so the current in the thin section also is 20.0 A . b) The resistance of the 3.00 In length of wire is given by Equation 19.13 on page 842 of the text, p13, (1.77 X 18—8 Q-m)(3.00 m) _3 A were x 10—3 m )/2}2 7 X C) The resistance of the :00 :11 length of wire is p2 . {1.77 x 10—8 Q-m)(1.00 In) 12—— ..._ I _3 . A W~225x10 e d) The two resistances are in series, so the total resistance is their sum, 3— R: R3 +131 = 7.51 x 10-3 n + 22.5 ><1O‘3 n a 30.0 x 10-3 £2. 9) The potential difference between the ends of the wire is found from Ohm’s iaw, v 2 re 2 (20.0 A ){300 x 10“3 n) = 0.600 V. 19.26 The Wire connecting the to Progressing from A to B: 59 109 109 59 The tWO pairs of 10 Q resistors are in paraHEI, product divided by the sum, USing this result, our circuit looks like AWVMW—AA/WNB 5Q . 5.0 9 5.0 .Q 5 9 19.31 Since the battery is connected direct} y to the motor, the motor and battery are simuitaneously in series and parallel with other: starter motor P = IV : (90.0 A)(12.D V) a 1.08 x 103 W = 1.08 kW. 40.77 x 10-5 o-m)(5.0 x 10-2 m) :2} d: 7r(5.0 x 10—5 Q/m)(5.0 X 10-2 m) 2:; 01:21 >< 104m. b) The potential difference is V = IR = (1.00 x 103 A)(5.0 >< 10*5—3x50 x104 m) z 2.5 x 10—3 V. c) The power absorbed by the wire segment is ry and 15.0 9 resistor er ore, they have the same potential diffs 15.0 D resistor is 30.0 V. rence across them. Hence the po 5) The power absorbed by the 15.0 D resistor is arrangement then looks like: 20.0 D.- The 20.0 9 resistor is in parallel with the source; hence the potential difference across it also is 30,(} V) and we can use Ohm’s law to ﬁnd the current through it. V=IR =:> 30.0V=I(20.0.Q) ::> I:‘1.50A. 20.0 (2 The power absorbed by the battery is the potential difference across it, P = (—3.50 A)(30.0 V) = —105 W. e) The current through each of the l 0.0 Q resistors is 1.50 A. The potential difference across each is found from Ohm’s law; for either resistor, V : IR 2 (1.50 A)(10.0 Q) = 15.0 V. The power absorbed by each is P 2 IV = (1.50 A){15.G V) = 22.5 W. 19.68 a) The potential energy stored in the capacitor is PE : 5-ch =¢ 1.00 J z %0(500 V)2 =¢ C' = 8.00 ><10“S F = 800,013. b) The time constant is ‘1’ 2 RC, so ﬁve time constants is 57 : 5R0 :> 0.100 x 10—3 s : 5R(8.(}0 x 10‘”6 F) a: R = 2.50 o. c) From Example 19.19 on page 876, the potential energy varies with time as The instantaneous power P is 1 1 5” me P 2 —— : -— -——-—— < > 57“ Gs Paw]: 57' OS dt d]: PE 5 PEG Hm PE; _5 PEG -5705 m—w—ST —1)_——5T (4.54x10 —1)_——-5T (—1) 1.00 J _———-.__.m_ = W200 103 W. 5(0100 x 10—3 s) X 80, as the capacitor bank discharges, it suppiies ah average power of 2.00 kW, or 2000 3/8. \ .- maynmff‘ﬂ’wlml‘ﬁw‘mf-ﬂﬂxw‘ "Man"....m.-_—:-1~,...w~._.,WM....,...,....,....,...,...,.....,...,...:.. .-, n-u.-.,ww:.._m..,. "Hiywﬂw , . _mm,,..w,.vﬂaa..wm_____.ww hum H .__,_.v.........-_...—_.,-_.wmw..a 19.7 2 a) The current in the branch containing the—capacitor is zero, since the capacitor acts as an open switch. {3) Since there is no current in the branch with the 1.0 k9 resistor, the potential dih’ererrce across it is 0 V A (from Ohm’s an). C) The circuit effectively is only a single loop, since the part between nodes A and B with the capacitor carries no Current. Earthermore, the resistors in the part of the circuit carrying current are in series, so they can be replaced by a single resistor with resistance 2.0 k-Q —: 3.0 k9 = 5.0 k0. To ﬁnd the current, apply the KVL clockwise around the loop, —2[}V +I(5.0x mammov : 1:4.0 X 10*»; =4.0mA._ d) Use Ohm’s law: V3.0 m : IR : (are xio-3 A)(3.0 x 103 c} = 12 V. e) Since there is 0 V across the 1.0 k9 resistor, the capacitor eﬁectively is in parallel with the 3.0 M2 resistor, and so also has 22 V across it. f) There is zero power absorbed by the capacitor and the 1.0 kﬂ resistor. The power absorbed by the 3.0 M} resistor is P30 m : 1V3.) m 2 (4,0 x 10-3 A )(12 V) m 4.8 x 10—2 W, The power absorbed by the 2.0 tr!) resistor is 7 PM m : IVM M; x we) = 121% = (4.0 x10“3 A)2(2.0 x 103 o) e 3.2 x 10"2 W. The power absorbed by the battery is Paw = (—4.0 x 10-3 A){20 V) 2 43.0 x 10—2 W, The total power absorbed by all circuit elements is Heta1=—8.UX10_2W +3.2x 10‘2W +4.8x10“2W+GW +ow :OW. ...
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## This note was uploaded on 10/01/2009 for the course PHYS 102 taught by Professor Henry during the Spring '09 term at Johns Hopkins.

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hw4 - 19.24. 599 19.24 a) The current in the wire is the...

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