Solucionario circuitos eléctricos - dorf, svoboda - 6ed

solucionario

1. EL SOLUCIONARIO
DATE TT A RI.

9. PH) Watts ce,

14. Y _ (0 cos 1)

17. vy and ade
P=vi=21

isthe power absorbed by the resistor

18. and i, adhe

and À

and 6, do not adhere to the passive convention so

The power absorbed by R, is

vis >90) = 16M

‘Model the heater as a resistor, then

witha 250 V sour

witha 210 V source CIO” 056 w

19. P5000 12:
The eurent required by the mine lights is 4
" sh mo 3
inthe wir
‘maximum resistance ofthe copper wire allowed is
= 005. 0055000 _ 1446
Pasa

now since the length of the wire is Z = 2:100 = 200 m= 20,000 em

us R=pL/A with p=1.7:10"0-cm fiom Table 25-1
PL _ 1.710 2
4 0.236 em’
R 0144

20. onsider the eurent sou
+, and v, do not adhere tothe passive convention,
12

supplied b

'onsider the voltage sowee:

+, and, do adhere tothe passive convention,
P= 4, 1312 = 36W,

is the power absorbed by the volt

The voltage source suppli

‘Consider the current soure
+, andy, adhere to the passive convention

0 P= 1, ¥,=312 = 36V

is the power absorbed by the current soure

Curent source supplies

onsider the voltage sours
and v, do not adhere to the passive convention
y= 3412-36 W

lag

23. 510

10 V and +, ‘A. (Notice that the ammeter me rather han 4, )

ect the resistor eurent tobe #= Ÿ 0.48 A. The power absorbed by

this resistor willbe P = # = (0.48) (I

A half watt resistor cant absorb this much power. You should not try another resistor.

31. Power absorbed by the 4

Power absorbed by the 8 © resistor

35. Voltage division

100

using voltage divider: »

with y, = 20V and yy > 9 V

with y, = 28V and yy < 13, R

72. vi

18vdc

151. ELITE TU)

153. un

161. 1
L{2ao*)(00) = 11

164. (ejer ] (ate

1
(are)
4

167. g-Cx(0)_15010*-(15x10°)(5)

(340*)eos(soors4s')= € À 1-45 )= C(12)(-500)sin(5001-45 )

C(6000)cos( 50045 )

173. wi!) and > #0) =0

»(1040°)=5-5 5 > w(lo)=1

since we), = 0 > > 040) sinfion?

(2a0*)(25)

First non-negative ¢ for max energy oocurs when: 10

174. on capacitor = Cv = (10:10) (6)

Ag _ 60:10
+> Todo

stored energy = 10

es with 3uF LESA our
uF uF

(60081001) = (2:10) (6) (100) (-sin100 A= =1.2 sintoOr mA

4 uP in series with 4 yr = HE Ey”
Hal uF

2 ur |2 FA ur

4 Pin series with 4 LP = 2 u

a
9-20 3e 2:10) (0+3(-250)

in series with 5 ¢

(25:10"0082501 = (Sc) qasin2500) 50605 2801

10 c 10x10 = 10 nF

175. Find max. voltage across coil: 200 [100(400)cos 4001]

8410 y

Vou = EX10% V thus have a field of #10

which exceeds dielectric strength in air of 3:10

We get a discharge as the airis ionized

Ate") +10(Ate") = 0 4e "+ 39.64

176. (0.25X0.12\300)e05(5001- 30")

[uo de

Las ¥,@=4n
op
42:07 de
no |
410
(bao
sagt HO")

(oo ie San’ 10 -140°5-$00*(-021-104)
va

5:10

340° | - 140° = -1.6 na

177. 110
octet us 40) = (1) 10%
! liao)"

w(t) = 4010!) e + 440°(1a0°)
for Las << us 4 (0=1 ma

0 = 40/40") + (4410

bao
110

for Zu 1 = 40
(2:10)(440°-10") +400

when $y <te 7us (D ==1:107 and

w(t) =(2«10°)(10) =-2 v

when Tys<t<8us «(=| M0 > = 110
bo a

10°1-8x10°)+(4x10°)(10°)=-12+(2x10°) 1
en Su <1,then,() = 0

178. v(r)=100«10

PORTO Ostet

wn

179. asin2s) asin29

80 [200524 sin24]

O sin(2r+20 + sin(2r-20)]= 80 sin dr W

sofsinar dr ® [cosarf]=20 1=cox40

.036[1- 0052001] J= 36 [1-00 2004 ] ml

Sm AS. Smit | Smit

and 4mH+ 4 mH =8mH

0 = (84107) À (54387) = (B10) (043250) e

180. \4:10)(250) cos 2501

183. 200k
H(t} A

20k0 7

TS

20k0
WOMAN

184. 200k Yur ma nur

FU Ma dur

185. 0.065 = 0.02

3
on

0.115 2.0065

0.065 =0.065 (Y

100