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Unformatted text preview: scan/0M VersionC
Ryerson University
Department of Electrical and Computer Engineering
Midterm Examination, EESSIZ Electric Circuits Duration: 1.5 hours
October 2009 Student’s Name: ....................................................................................... Student Number: ......................................... Section: ....................... Professors: D Prof. Ghorab [:1 Prof. Karim
D Prof. Hussain
D Prof. Umapathy NOTES: 0 You must indicate your section, and tick mark your professor’s name above. This is a closed book examination. Some useful equations are given in the last page.
You are encouraged to show the steps of your calculations. Part marks are given for
correct steps even for wrong ﬁnal results. NO QUESTIONS to be asked. If doubt exists as to the interpretation of any question, you are urged to submit with the answer, a clear statement of any logical assumptions made. Answer all questions. The questions have equal marks.
Mark obtained Q1:
Parta: A 2—hp motor runs on 120V supply with 80% 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 kilowatthour. [3x2] Partb: 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]
[max 100 a, 0.25 W
‘———’
R1
Vmax R2 200 :2, 0.125 W
Figure—1 Eggs; A power source can supply a load with 1.2 kWh of energy in 40 minutes. If the current
drawn is 6A, 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 1.08 MJ has been supplied by the source with the same current and voltage. [3x2] Q1
a)
Solution
(04%): Pout x100
Pin
0.8: ”7464593 (12112274611)
an Pin
gn=t422=1865 W @
0.8
Li?)£—— I=fii=1§6i=1554 A @
V 120
@945 P —1865 —1 865 kW
' * "" _ 1000 _ '
W (kWh) =P(kW)><t (h) =1.865x10 =18.65 kWh
cost =18.65x8 = 149.2 Cents = $1.492 @
b)
Solution ﬂmax 0.25
1 = l = "—— =0.05 A
lmax RI 100 15m“ 0.125
1 = f =‘/—— =0.025 A
2m” R2 200 Clearly, the maximum safe current is: Im=0. 025 A @ 4’ 1 Thus R2 is the weakest link in this connection. Hence, Vmax = ImaXRT = 1max (R1 + R2) = 0.025000 + 200) = 0.025 x 300 = 7.5 V @w @_ PIE/1:: 1.2x1000x3600 __
2‘ 40x60 1.08x106 J=300x6xt t_1.08><106
1800 = 600 s 2 10min —1800 W @ E/ 3“ (ix/“Q "2a {A
“(L/ 935$)?“ \j M Q2:
For the circuit shown in Figure2, calculate V1, V2 , 11,12 , I3 and I4 . [6+6+2+2+2+2] , v2§ .  
10A ggouagoszf—P a, ..
«= 41"" l/Qfg‘ﬂw ~ C Q3: In the Figure3, E = 20V, R1 = 5m, R2 = 81d), R3 = 6k!) and R4 = lOkﬂ (i) Calculate the voltage Vx , Vy and ny . [12]
(ii) If ny = 0, and R1 = lkﬂ, R2 = 3190 and R3 = 2kg, calculate value of R4.
Also calculate the currents through R1 and R4 under this condition. [4+2+2] Q4: m 29?“. M For the circuit shown in Figure4, (a) calculate the parameters (R31 and V3,) of the Thevenin’s equivalent circuit at terminals a — b, as seen by the load RL. [5+5]
(1)) Determine the value of RL for maximum power transfer to RL [4]
(c) Calculate the maximum power which can be delivered to RL . [6]
1D 2. 5 Q 2. 5 V
at
20 hi
2.5 A 1“ 4.5V— F i gure—4 @4194 , , , ”VafS.bm(c \ 9x) , ,/ ”if FLA Cl, [[751 ‘ {/g 3179,44 gﬂmqg, ,i‘, , FarAwfw i: r SQs—IM ”20‘ ”1””: °‘ Some useful gquations I=—Q— V=R.I lhp = 746 Watts V2 Watt time(sec) P=V.I=IZ.R=—— 1kwh=—————x
R 1000 10 3600 ...
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 Winter '11
 Karim

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