rizzoni 6th solution pdf ch.02 PDF

Preview

Summary

Download rizzoni 6th solution pdf ch.02 PDF

Description

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2

Problem 2.26 If an electric heater requires 23 A at 110 V, determine a. The power it dissipates as heat or other losses. b. The energy dissipated by the heater in a 24-hperiod. c. The cost of the energy if the power company charges at the rate 6 cents/kWh.

Solution: Known quantities: Cu r r e n ta bs or b e db yt h ehe a t e r ;v o l t a g ea twhi c ht hec ur r e n ti ss up pl i e d;c o s toft hee ne r g y .

Find: a ) Po we rc on s umpt i on b) En e r g ydi s s i p a t e di n2 4hr . c ) Co s toft h eEn e r gy

Assumptions: Th ehe a t e rwo r ksf o r24h ou r sc o nt i nu ous l y .

Analysis:

P=VI =110 V (23 A )=2.53×103 W =2.53 KW J s W=Pt=2 .53×103 ×24 hr×3600 =218 . 6 MJ=60 . 72 KW-hr s hr b ) $ Cost=( Rate )×W =0 . 06  ( 2 .53 kW)  (24 hr ) =$ 3 . 64 kW−hr c ) Problem 2.27 a )

In the circuit shown in Figure P2.27, determine the terminal voltage of the source, the power supplied to the circuit (or load), and the efficiency of the circuit. Assume that the only loss is due to the internal resistance of the source. Efficiency is defined as the ratio of load power to source power.

Solution: Known quantities:

v =12 V Ci r c ui ts h owni nFi gu r eP2 . 27wi t hv o l t a g es ou r c e , s a ndr e s i s t a nc eoft hel o a d ,

R0 =7 k 

;i n t e r na lr e s i s t a n c eoft h es o ur c e , Rs =5 k  ;

.

Find: Th et e r mi na lv ol t a g eoft h es ou r c e ;t hep o we rs up pl i e dt ot h ec i r c ui t ,t hee ffic i e nc yo ft hec i r c ui t .

Assumptions: As s u met ha tt heo nl yl os si sd uet ot h ei n t e r n a lr e s i s t a nc eoft hes our c e .

2. 2 1 PROPRI ETARY MATERI AL.© TheMc Gr a wHi l lCo mpa ni e s ,I n c . Li mi t e ddi s t r i bu t i onp e r mi t t e don l yt ot e a c h e r sa nd e d u c a t o r sf orc ou r s ep r e pa r a t i o n.I fy o ua r eas t ud e n tus i n gt h i sMa nu a l , y o ua r eu s i n gi twi t ho u tpe r mi s s i o n .

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2

Analysis:

KVL : −v S +i T RS + vT =0 OL: v T =i T R0 vT ∴ iT = RL vT −v S + RS +v T =0 R0 vS 12 V vT = = =7 V or 5 k RS 1+ 1+ 7 k R0 v S R 0 12 V⋅7 k  = =7 V . VD : v T = R S + R 0 5 k +7 k  2 2 vT (7 V ) =7 mW PL = = R0 V 7×103 A 2 PL Pout i R0 7 k =0 . 5833 or 58. 33 % = η= =2 T 2 = P in P R + P L i T R S +iT R 0 5 k +7 k  S

.

2. 2 2 PROPRI ETARY MATERI AL.© TheMc Gr a wHi l lCo mpa ni e s ,I n c . Li mi t e ddi s t r i bu t i onp e r mi t t e don l yt ot e a c h e r sa nd e d u c a t o r sf orc ou r s ep r e pa r a t i o n.I fy o ua r eas t ud e n tus i n gt h i sMa nu a l , y o ua r eu s i n gi twi t ho u tpe r mi s s i o n .

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2

Problem 2.28 A 24-volt automotive battery is connected to two headlights, such that the two loads are in parallel; each of the headlights is intended to be a 75-W load, however, a 100-W headlight is mistakenly installed. What is the resistance of each headlight, and what is the total resistance seen by the battery?

Solution: Known quantities: He a dl i gh t sc o nne c t e di np a r a l l e lt oa2 4Va u t omot i v eba t t e r y ;p o we ra b s o r be db ye a c hh e a d l i g ht .

Find: Re s i s t a nc eo fe a c hhe a d l i ght ;t ot a lr e s i s t a n c es e e nb yt h eb a t t e r y .

Analysis: He a dl i gh tn o .1 :

P=v ´ i=100 W = R=

v2 or R

v2 576 =5 .76  = 100 100

He a dl i gh tn o .2 :

P=v ´ i=75 W = R=

v 2 or R

v 2 576 =7 . 68 = 75 75

Th et ot a lr e s i s t a n c ei sg i v e nb yt hep a r a l l e lc o mbi na t i on : 1 1 1 = + R TOTAL 5 .76  7 .68  orR 2 9 TOTAL=3.

Problem 2.29 What is the equivalent resistance seen by the battery of Problem 2.28 if two 15-W taillights are added (in parallel) to the two 75-W (each) headlights?

Solution: Known quantities: He a dl i gh t sa n d24 Va ut omo t i v eb a t t e r yo fpr o bl e m2. 13wi t h21 5Wt a i l l i gh t sa d de di npa r a l l e l ;po we ra bs or b e db y e a c hhe a dl i gh t ;p o we ra b s or b e db ye a c ht a i l l i ght .

Find: Eq ui v a l e ntr e s i s t a n c es e e nb yt heba t t e r y .

Analysis: Th er e s i s t a n c ec o r r e s p ond i n gt oa7 5Wh e a d l i g hti s : 2. 2 3 PROPRI ETARY MATERI AL.© TheMc Gr a wHi l lCo mpa ni e s ,I n c . Li mi t e ddi s t r i bu t i onp e r mi t t e don l yt ot e a c h e r sa nd e d u c a t o r sf orc ou r s ep r e pa r a t i o n.I fy o ua r eas t ud e n tus i n gt h i sMa nu a l , y o ua r eu s i n gi twi t ho u tpe r mi s s i o n .

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2 2 v 576 =7 . 68 R75 W = = 75 75

Fo re a c h15 Wt a i ll i g htwec omp ut et h er e s i s t a nc e : 2

R 15 W =

v 576 =38 . 4  = 15 15

Th e r e f o r e ,t het o t a lr e s i s t a nc ei sc ompu t e da s : 1 1 + 1 + 1 + 1 = R TOTAL 7 . 68 7 .68  38 . 4  38 . 4  orRTOTAL=3 . 2

Problem 2.30 Using the circuit in Figure P2.30, determine the power absorbed by the potentiometer and plot it as a function of R

Solution: Known quantities: v =15V, Rs =10Oh ms ,a n dt hec i r c ui ti nFi g ur eP2. 30 . s

Find: R

Analysis: Us eo hmsl a wt ofinda ne qu a t i onf orPa saf u nc t i ono fR:

PR =V R∗I R Th ev ol t a g ea c r os sRi se q ua lt ot hes ou r c ev o l t a g emi n ust hev o l t a g ea c r os sRs :

V R =15 V −V Rs

VRSi sd e t e r mi ne db yt hec ur r e n tt h r oug ht hel oo pwh i c hc a nbef ou ndb ya ddi n gt her e s i s t or si ns e r i e s :

15V (R s + R) V RS =10 Ω∗I R I R=

Si mp l i f y :

[ (

PR = 15 V − 10

)] [

Ω∗15 V 15 V ∗ 10 Ω+ R 10 Ω+ R

]

Pl o t :

PR

PROPRI ETARY MATERI AL.© TheMc Gr a wHi l lCo mpa ni e s ,I n c . Li mi t e ddi s t r i bu t i onp e r mi t t e don l yt ot e a c h e r sa nd e d u c a t o r sf orc ou r s ep r e pa r a t i o n.I fy o ua r eas t ud e n tus i n gt h i sMa nu a l , y o ua r eu s i n gi twi t ho u tpe r mi s s i o n .

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2

R

Problem 2.31 Using the circuit in Figure P2.27 to determine the following values relating to the power supply and consumption of the

Solution: Known quantities: v =15V, Rs =10 0Oh ms , i ,1 0, 2 0,30 ,8 0,10 0mA. s T=0 Th ec i r c u i ti nFi g ur eP2 . 2 7.

Find: a ) Th et ot a lpo we rs up pl i e db yt hei d e a ls o ur c e b) Th epo we rdi s s i pa t e dwi t hi nt henon i de a ls ou r c e c ) Ho wmuc hpo we ri ss up pl i e dt ot h el oa dr e s i s t o r d ) Pl o tv ndpo we rs u pp l i e dt oR0a saf u nc t i ono fi Ta T.

Analysis: a ) Th epo we rs up pl i e db yt hei de a ls our c ei se qua lt ot h ec ur r e n tt h r ou ght hel oo pt i me st he1 5Voft h es u pp l y . Fr om c u r r e ntl o we s tt ohi g he s tt h epo we rs u ppl i e dwo ul db e :

0 W 0.15 W 0.3W 0.45 W 1.2 W 1.5 W b) Th epo we rdi s s i pa t e dwi t hi nt henon i de a ls ou r c ei st hepo we rd i s s i p a t e db yRswhi c hc a nb ef oun dus i n g 2 P=i *r .Fr omc ur r e ntl o we s tt oh i ghe s tt hepo we rdi s s i pa t e dwo ul db e :

0 W 0.01 W 0.04 W 0.09 W 0.64 W 1 W c ) Th epo we rs up pl i e dt ot h el oa dr e s i s t ori se qu a lt ot h et ot a lpo we rs upp l i e dmi nust hep o we rdi s s i pa t e db y t h eno ni d e a ls o ur c e .Fr om c ur r e n tl o we s tt oh i gh e s tt hep o we rs upp l i e dwou l dbe :

0 W 0.14 W 0.26 W 0.36 W 0.56 W 0.5 W d ) Fo rt h evT pl otOhm’ sLa wc a nbeu s e dt ofin dt hev ol t a g edr o pa c r o s s whi c hi s v 5–1 00i Fort he sR T =1 T. po we rpl o t ,t hed a t af r omp a r tcc a nbeu s e ddi r e c t l y .

2. 2 5 PROPRI ETARY MATERI AL.© TheMc Gr a wHi l lCo mpa ni e s ,I n c . Li mi t e ddi s t r i bu t i onp e r mi t t e don l yt ot e a c h e r sa nd e d u c a t o r sf orc ou r s ep r e pa r a t i o n.I fy o ua r eas t ud e n tus i n gt h i sMa nu a l , y o ua r eu s i n gi twi t ho u tpe r mi s s i o n .

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2

Vs across R2 in figure P2.32 should be replaced with V2 Problem 2.32 In the circuit in Figure P2.32, assume v2 = vs/6 and the power delivered by the source is 150 mW. Find R, vs, v2, and i.

Solution: Known quantities: R1=8k Ω,R2=1 0kΩ,R3=12k Ω,P 5 0mW a ndt h ec i r c ui ti nFi g ur eP2 . 32. s=1

Find: R,vs, v2, and i.

Analysis: Us eo hmsl a wt ofind a n di : sv

2. 2 6 PROPRI ETARY MATERI AL.© TheMc Gr a wHi l lCo mpa ni e s ,I n c . Li mi t e ddi s t r i bu t i onp e r mi t t e don l yt ot e a c h e r sa nd e d u c a t o r sf orc ou r s ep r e pa r a t i o n.I fy o ua r eas t ud e n tus i n gt h i sMa nu a l , y o ua r eu s i n gi twi t ho u tpe r mi s s i o n .

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2

v v 2= s =R2∗i 6 Al s o:

v s∗i=150 mW So :

√

i= 2

150 mW =1.58 mA R2∗6

Si n c ewekn o wi ,vsc a ne a s i l ybef ou nd :

v s∗i=150 mW v s=94.94 V

v 2=

vs =15.82V 6

Us eOh m’ sl a wt ofin dRe q:

vs =Req =60 k Ω i Us eRe qt ofindR:

Req =R1 + R+ R2 + R3 R R=R eq−(¿ ¿ 1+ R2 + R3 )=30 k Ω ¿

Problem 2.33 A GE SoftWhiteLonglife lightbulb is rated as follows: PR = rated power = 60 W POR = rated optical power = 820 lumens (lm) (average) 1 lumen = 1/680W Operating life = 1,500 h (average) VR = rated operating voltage = 115 V The resistance of the filament of the bulb, measured with a standard multimeter, is 16.7 Ω. When the bulb is connected into a circuit and is operating at the rated values given above, determine a. The resistance of the filament. b. The efficiency of the bulb.

Solution: Known quantities: Ra t e dp o we r ;r a t e dopt i c a lp o we r ;ope r a t i n gl i f e ;r a t e do pe r a t i n gv ol t a g e ;o pe n c i r c u i tr e s i s t a n c eo ft hefil a me nt .

2. 2 7 PROPRI ETARY MATERI AL.© TheMc Gr a wHi l lCo mpa ni e s ,I n c . Li mi t e ddi s t r i bu t i onp e r mi t t e don l yt ot e a c h e r sa nd e d u c a t o r sf orc ou r s ep r e pa r a t i o n.I fy o ua r eas t ud e n tus i n gt h i sMa nu a l , y o ua r eu s i n gi twi t ho u tpe r mi s s i o n .

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2

Find: a ) Th er e s i s t a n c eo ft hefil a me nti nop e r a t i on b) Th ee ffic i e nc yo ft heb ul b .

Analysis: a )

P=VI ∴ I = R= OL:

P R 60 W =521 .7 mA = V R 115 V

V V R 115 V =220 .4  = = I 521 .7 mA I

b ) Effic i e n c yi sde fin e da st h er a t i ooft h eus e f u lp o we rdi s s i pa t e db yors u pp l i e db yt hel oa dt ot h et ot a lpo we rs u pp l i e d b yt h es o ur c e .I nt hi sc a s e ,t h eus e f u lpo we rs u ppl i e db yt h el oa di st heop t i c a lpo we r .

Po,out =

=820 lum Op t i c a lPo we rOut

=

W =1. 206 W 680 lum

Po , out 1. 206 W = =0 . 0201=2 . 01 % PR 60 W

η= effici e nc y

.

Problem 2.34 An incandescent lightbulb rated at 100 W will dissipate 100 W as heat and light when connected across a 110-V ideal voltage source. If three of these bulbs are connected in series across the same source, determine the power each bulb will dissipate.

Solution: Known quantities: Ra t e dp o we r ;r a t e dv o l t a g eofal i g htbu l b .

Find: Th epo we rd i s s i p a t e db yas e r i e soft h r e el i ghtb ul b sc onn e c t e dt ot henomi n a lv o l t a g e .

Assumptions: Th er e s i s t a n c eofe a c hb ul bdo e s n ’ tv a r ywhe nc o nne c t e di ns e r i e s .

Analysis: Whe nc o nne c t e di ns e r i e s , t hev ol t a g eo ft hes ou r c ewi l ldi vi d ee qu a l l ya c r os st het h r e eb ul bs . Thev ol t a g ea c r os s e a c hbu l bwi l lbe1/ 3wha ti twa swhe nt heb ul bswe r ec o nne c t e di n di v i dua l l ya c r o s st h es o ur c e .Po we rd i s s i p a t e di n ar e s i s t a nc ei saf u nc t i o no ft h ev ol t a g es qua r e d ,s ot hep o we rdi s s i pa t e di ne a c hb ul bwh e nc onn e c t e di ns e r i e swi l l b e1/ 9wh a ti twa swh e nt hebul b swe r ec onn e c t e di ndi vi d ua l l y ,or11 . 11W:

2. 2 8 PROPRI ETARY MATERI AL.© TheMc Gr a wHi l lCo mpa ni e s ,I n c . Li mi t e ddi s t r i bu t i onp e r mi t t e don l yt ot e a c h e r sa nd e d u c a t o r sf orc ou r s ep r e pa r a t i o n.I fy o ua r eas t ud e n tus i n gt h i sMa nu a l , y o ua r eu s i n gi twi t ho u tpe r mi s s i o n .

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition

Ohm’ sLa w:

Problem solutions, Chapter 2 V 2B P=IV B =I 2 RB = RB V B = V S =110 V

V 2B ( 110 V )2 =121  RB = = P 100 W Con ne c t e di ns e r i e sa n da s s u mi ngt her e s i s t a nc eofe a c hb ul br e ma i nst hes a mea swh e nc onn e c t e di ndi vi du a l l y: − V S +V B1 +V B 2 +V B 3 = 0 −V S + IR B 3 + IR B 2 + IR B 1=0 OL: KVL:

I=

VS

110 V =303 mA = R B 1 + R B 2 + R B 3 ( 121+ 121+ 121 )

1 PB 1 =I 2 R B 1 =( 303 mA)2 ( 121 )=11 . 11 W = 100 W . 9

Problem 2.35 An incandescent lightbulb rated at 60 W will dissipate 60 W as heat and light when connected across a 100-V ideal voltage source. A 100-W bulb will dissipate 100 W when connected across the same source. If the bulbs are connected in series across the same source, determine the power that either one of the two bulbs will dissipate.

Solution: Known quantities: Ra t e dp o we ra ndr a t e dv o l t a g eo ft het wol i g htbu l bs .

Find: Th epo we rd i s s i p a t e db yt h es e r i e so ft het wol i g htbu l b s .

Assumptions: Th er e s i s t a n c eofe a c hb ul bdo e s n ’ tv a r ywhe nc o nne c t e di ns e r i e s .

Analysis: 2

P=IV B =I 2 RB = Ohm’ sLa w: V B = V S =110 V

R60=

VB RB

V 2B (110 V ) 2 = =201 .7  P60 60 W 2

V B ( 110 V ) 2 =121  = R100= P100 100 W Whe nc o nne c t e di ns e r i e sa nda s s umi ngt h er e s i s t a nc eofe a c hb ul br e ma i nst hes a mea swh e nc on ne c t e d i n di v i dua l l y : −V S + V B 60+ V B 100=0 KVL: −V S + IR B 60 + IRB 100 =0 OL: 2. 2 9 PROPRI ETARY MATERI AL.© TheMc Gr a wHi l lCo mpa ni e s ,I n c . Li mi t e ddi s t r i bu t i onp e r mi t t e don l yt ot e a c h e r sa nd e d u c a t o r sf orc ou r s ep r e pa r a t i o n.I fy o ua r eas t ud e n tus i n gt h i sMa nu a l , y o ua r eu s i n gi twi t ho u tpe r mi s s i o n .

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2 I=

VS R B 60 + RB 100

110 V

= =340 . 9 mA V 201 . 7+121 A 2

PB 60=I 2 R B 60=( 340 .9 mA ) ( 201. 7  )=23. 44 W 2 PB 100 =I 2 R B 100=( 340 . 9 mA ) ( 121  )=14 .06 W No t e s :1. I t ’ ss t r a n g ebu ti t ’ st r u et ha ta60W b ul bc o nne c t e di ns e r i e swi t ha100W b ul bwi l ld i s s i p a t emor epo we r t h a nt he1 00W bu l b .2 .I ft hepo we rdi s s i pa t e db yt hefil a me n ti nab ul bd e c r e a s e s ,t het e mp e r a t u r ea twh i c ht he fil a me n to pe r a t e sa ndt h e r e f o r ei t sr e s i s t a nc ewi l lde c r e a s e .Thi sma det h ea s s ump t i o na b outt her e s i s t a n c ene c e s s a r y .

Problem 2.36 For the circuit shown in Figure P2.36, find the voltage vab and the power dissipated in R2.

Solution: Known quantities: Circuit of Figure P2.36. R1=5Ω, R2=3Ω, R3=4Ω, R4=5Ω, vs=12V. Find: vab and the power dissipated in R2.

Analysis: Fi n dt hec ur r e n tt h r ou ghe a c hr e s i s t orpa i rb yc o mbi n i n gt he mi ns e r i e s :

R1 + R2=8 Ω R3 + R4 =9 Ω

Si n c et he r ei s1 2Va c r os sb ot ho ft he mOh m’ sl a wi su s e dt ofin dt hec ur r e nt s :

12V =8 Ω∗I a I a=1.5 A 12V =9 Ω∗I b I b=1.33 A No wfin dt hev o l t a g edr o pa c r os s a ndR3: 1R

V a=12 V −(I a∗R1 )=4.5V V b=12 V −(I b∗R3 )=6.67 V V ab=−2.17 V P se q ua lt oI i me sVa: R2i bt

PR 2 =4.5 V∗1.5 A=6.75 W

Problem 2.37 For the circuit shown in Figure P2.37, find the currents i2 and i1 and the power supplied by the voltage source.

2. 3 0 PROPRI ETARY MATERI AL.© TheMc Gr a wHi l lCo mpa ni e s ,I n c e d u c a t o r sf orc ou r s ep r e pa r a t i o n.I fy o ua r eas t ud e n tus i n gt h i sMa nu a l ,

t e a c h e r sa nd

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 2

Solution: Known quantities: Circuit of Figure P2.37. R1=20Ω, R2=12Ω, R3=10Ω, Is=3A, vs=7V. Find: i1, i2, and Pv.

Analysis: Us eKVLoft hel e f t mos tl o op :

v s−R1∗i1−R 2∗i 2−R3∗i 1=0 v s=R1∗i1 +R2∗i 2 +R3∗i1 Us eKCLwh e r ei e t si 1 me s:

i1+ is −i2=0

i1+i s =i2 Comb i net het woe q ua t i on st os ol v ef ori : 1

v s=R1∗i1 +R2∗( i 1+is)+ R3∗i1 i1=−0.69 A So l v ef o ri 2:

i 1+ i s = i 2 i 2=2.31 A Po we rde l i v e r e de qua l st hev i me si : st 1

Pv =v s∗i1=−4.83 W

Problem 2.38 For the circuit shown in Figure P2.38, find the currents i2 and i1 and the power supplied by the voltage sources.

Solution: Known quantities: Circuit of Figure P2.38. R1=18Ω, R2=10Ω, v1=15V, v2=6V. Find: i1, i2, Pv1,Pv2.

Analysis: Us eKVLoft her i gh t mo s tl oop :

v 2 +R1∗(−i1)− R2∗i 2=0 v 2=R 1∗i1+ R 2∗i2 Us eKCLwh e r ei e t si 1 me s: 2. 3 1 PROPRI ETARY MATERI AL.© TheMc Gr a wHi l lCo mpa ni e s ,I n c . Li mi t e ddi s t r i bu t i onp e r mi t t e don l yt ot e a c h e r sa nd e d u c a t o r sf orc ou r s ep r e pa r a t i o n.I fy o ua r eas t ud e n tus i n gt h i sMa nu a l , y o ua r eu s i n gi twi t ho u tpe r mi s s i o n .

Full file at https://testbanku.eu/G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition

i−i1−i2=0

Problem solutions, Chapter 2

i1+ i2=i Th ea s s ump t i o nc a nb ema d et h a tv se q ua lt ot h ev ol t a g ea c r os sR2. 1i

v 1=R 2∗i 2 i2=1.5 A Th ea s s ump t i o nc a nb ema d et h a tt hev ol t a g edr opa c r o s sR1i se q ua lt ot hed i ffe r e n c ebe t we e n v ndv2. 1a

v 1−v 2=R1∗i1 i1=0.5 A So l v ef o ri :

i1+ i2=i i=2.0 A Po we rde l i v e r e de qua l st hev ol t a g es o ur c et i me st hec ur r e ntt h r o ughi t :

P1=v1∗i=30.0 W P2=v 2∗(−i1 )=−3 W

Problem 2.39 Consider NiMH hobbyist batteries shown in the circuit of Figure P2.39. a. If V1 = 12.0 V, R1 = 0.15 Ω and RL = 2.55Ωfind the load...