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Unformatted text preview: spAUT/o/V Version—B
Ryerson University
Department of Electrical and Computer Engineering Midterm Examination, EESSlZ Electric Circuits Duration: 1.5 hours
October 2009 Student’s Name: ..................................................................................... .. Student Number: ....................................... .. Section: ..................... .. Professors: [I Prof. Ghorab
D Prof. Karim
D Prof. Hussain
(2] Prof. Umapathy
NOTES: 0 You must indicate your section, and tick mark xour professor’s name above.
0 This is a closed book examination. Some useful guations are g_lg' en in the last page. 0 You are encouraged to Show the steps of your calculations. Part marks are given for correct steps even for wrong ﬁnal results. 0 NO QUESTIONS to be asked. Ifdoubtexists as to the interpretation of any question, you are urged to submit with the answer, a clear statement of any logical assumptions made. 0 Answer all questions. The questions have equal marks. Mark Q4 20 Total 80 Qt:
Parta: A 1.5hp motor runs on 120V supply with 85% efﬁciency, calculate (i) The input power delivered by the source.
(ii) The current drawn by the motor.
(iii) The cost of leaving the motor running for 10 hours, when the cost of electricity is
8 cents per kilowatt—hour. [3x2] 1 Z; Part—b: Calculate the maximum safe voltage and maximum safe current that can be applied to the two parallel resistors, whose values and maximum power ratings are shown in Figure1.
[2x4] : 0Q F igure—l Partc: A power source can supply a load with 1.8 kWh of energy in 40 minutes. If the current drawn is 9A, calculate (i) The power delivered to the load. (ii) The voltage across the resistors terminals. (iii) The time in minutes required if a total energy of 3.24 MJ has been supplied by the source with the same current and voltage. [3x2] : é. ,7 “WM—AV. 3.“...x,w.~m..~;wmw 01
a)
Solution « P
(1% 71%): out x100
Pin 1.5x746: 1119 (Mama
Pin Pz‘n 3,2222%“ W @
0.85 @1922— 1=5ﬂ=131647=1097 A (7:) V 120 0.85: Qt; 1316.47
, a: Pm=W=L31647 kW W(kWh)=P(kW)xt (h)=1.31647><10=13.1647 kWh
cost=13.1647><8=105.32 Cents=$l.05 @ b) M! Solution Vlmax =,/I1R1 = 0.25x100 =5 V Vzmax = /P2Rl = J0.36x225 = 9 V / ~\
3 k 0)
Clearly, the maximum safe voltage is: Vm=5 V Thus R1 is the weakest link in this connection. Hence, 1max = Vmax + Vmax =—5—+—5—= 0.0722 A  R1 R2 100 225 3 J1
(21 c) A W 1.8x1000x3600_ 
@} P=—t—=—W——27OO W @
@ﬂ¢~.P:Vx1
Vzﬁzglqumo V @
I 9 (ﬁOg~IV:thJUxt
324xuf.l=300x9xz _324fo
2700 t 21200 S=20min ‘2 {E if x ‘: «a? my g Maw” Q2:
For the circuit shown in Figure2, calculate V1, V2 , [1 , 12,13 and I4 . [6+6+2+2+2+2] 30 £3 Q3: (12') (1/) v\,;\/X..vy: (2') \/ :Exe. In the Figure—3, E = 15V, R1 = 3kﬂ, R2 = 6162, R3 = 4k!) and R4 = 9k!) (i) Calculate the voltage Vx , V3/ and ny . (ii) If ny = 0, and R1 = 2kQ, R2 = 4k!) and R3 2 4m, calculate value of R4.
[4+2+2] [12] Also calculate the currents through R1 and R4 under this condition. R3 R4 Figure3 (Pl+2231/5x *4 :@ V 2 (3+6)
YjE“ Q _:/5 .9 h/57 ,' '
(55*?3 yigjfa  73342 / IOm6&§;Lv655V “22' ‘ v
U vxyza, 1% n a; A/u///bgm% (9 ﬁZVi—Ti :V...
a; 1; ﬂgﬂ/i 4% / P2 057522) Q+£Dl03 l :]
f r f 3 5 __~ /5 m' l «3 w )
£3 £9 6% waff— LéHg/OB .. / 25x12) ﬂ,/ 025m ,3 Q4: For the circuit shown in Figure4, (a) calculate the parameters (R31 and V3,) of the Thevenin’s equivalent circuit at terminals 21 — b, as seen by the load RL. [5+5]
(b) Determine the value of RL for maximum power transfer to KL [4]
(c) Calculate the maximum power which can be delivered to RL . [6] 49 10.0.. 40V
———T———’Ws, I
$89
10A J— 72v; Figure4 , ,,,,,, i ,, *9 AW/ <,_.,,....,A,,,,..,,,, ,g,,,<,,_wg,,,,,h, #ﬂwhm, if,» “WWW” ,,,,,,,, ..... , W, ,,  , ,,,,,,,.,.‘ m “7.41 JLi ﬁv , .... ,,,,,,,,,,, ,j ,,,,,,, ,,,,,,, , ‘
Z ,ﬁﬂé icy/aauwﬁxlz , ,,,,,,,,,,,,,,,,,,,,,,, 7, r ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, ............. ‘_u ,,,,,,,,,,,,, ,7, ,,,,,,,,,,,,,,,,,,,,,,,,, #7 W V, r ,,,,,,,,,,,,,,,,,,,,,,,,,, . ,,,, w .... , ,7 ,,,,,,,,,,,,,,,,,,,,,,, )7 r. ' " " ’ , 5/
i r , 2 A ,1 ~ ' a ,, , a {j J Some useful gguations 1:9— V=R.I 2
V lkwh : Watt R 1000 10 lhp = 746 Watts time sec
x ( > 3600 ...
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 Winter '11
 Karim

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