This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 Qm)(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 Qm)(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 103 n + 22.5 ><1O‘3 n a 30.0 x 103 £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 105 om)(5.0 x 102 m)
:2} d:
7r(5.0 x 10—5 Q/m)(5.0 X 102 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 : 5ch =¢ 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‘mfﬂﬂxw‘ "Man"....m._—:1~,...w~._.,WM....,...,....,....,...,...,.....,...,...:.. ., nu..,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 kQ —: 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 xio3 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 103 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 103 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. ...
View
Full
Document
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.
 Spring '09
 HENRY
 Physics

Click to edit the document details