ECE 159S – Electric Circuit Fundamentals
3aLast Name:
First Name:
March 3, 2011
Student Number:
Your Tutorial Section (CIRCLE ONE):
01 (BA2139 W11-12)
02 (BA2135 W11-12)
03 (BA2185 W11-12)
04(BA2145 W11-12)
05 (BA2175 F1-2)
06 (BA2159 W1-2)
07(BA2159 F11-12)
08(BA2159 W10-11)
09 (WB144 W10-11)
10(BA2139 W10-11)
(YOU LOSE ONE POINT FOR INCORRECT TUTORIAL SECTION INFORMATION)
Test 2 – Answer All Questions
60 minutes
(Non-programmable Calculators Allowed)
Question 1 [10 points]
For the circuit shown in Fig. 1:
Fig. 1: Circuit for Question 1
a) Find the open-circuit circuit voltage between terminals A and B. [3 points]
VO.C.A-B = 10.6667 V
KCL at node M: 2+Ix+2Ix=0Ix=-0.6667 mA [1 point]
VO.C.A-B = 12+VR [1 point]
VR=IRx1 k
But from the bottom mesh: IR=2Ix =-1.333 mA[0.5 point]
Thus
VO.C. A-B = 12-1.333 = 10.6667 V [0.5 point]
b) Find the short-circuit circuit current between terminals A and B. [3 point]
KCL at node M: 2+Ix+2Ix=0Ix=-0.6667 mA [0.5 point]
IS.C.A-B = 10.6667 mA
KCL at Node B (inside the hashed box)
2Ix=IR + IS.C.A-B [1 point]
But IR=VR/1 k
Where by KVL in outer loop 12+VR=0  VR=-12 V
Thus IR-12 mA [1 point]
Then
IS.C.A-B = 2x-0.6667 – (-12) = 10.6667 mA [0.5 point]
c) Find the value of RL for the maximum power transfer condition.
[2 points]
For maximum power transfer  RL = RTh = VO.C.A-B / IS.C.A-B [1 point]
= 1 k[1 point]
d) If RL = 4 k, find the voltage Vo [2 points]
RL = 1 k
Using the Thevenin equivalent of the circuit, Vo is calculated using the voltage divider rule:
Vo = VTh x RL / (RL+ RTh) [ 1 point]
Vo = 8.53333 V
= 8.53333 V [ 1 point]
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ECE 159S – Electric Circuit Fundamentals
March 3, 2011
Question 2 [10 points]
a) The OP-AMP of the circuit shown in Fig. 2 is ideal. Find the relation between Vo, Vs1, and Vs2 [3
points]
d(Vo)/dt = d(Vs2)/dt +
1000(Vs2- Vs1)
Since the input current to the ideal OP-AMP is zero, then the following KCL equation holds:
IR = IC
IR = (Vs1- VN) / 10 kpoint]
Ic = 0.1F x d(VN-Vo)/dt point]
But since the OP-AMP is ideal, then
VN=Vs2 point]
Thus
d(Vo)/dt = d(Vs2)/dt + (Vs2- Vs1)/( 10 kx0.1F) point]
d(Vo)/dt = d(Vs2)/dt + 1000(Vs2- Vs1)
b) For the circuit shown in Fig. 3, the switch was open for a very long time then closed at t=0. Find an
expression for the capacitor voltage Vc(t). Sketch Vc(t) versus time. [7 points]
Vc(t)= 27 + 9 e(-6.667t) V
STEP1: find Vc(0)
Vc(0) = 36 V [ 1 point]
STEP4: find TC
TC=CRT=1.5 kxF
0.15 sec. [0.5 point]
STEP2: find Vc()
Vc()=36x6/(6+2)=27 V [2 point]
=
Fig. 3: Circuit for Question 2.b
STEP3: find RT
RT=2//6 k = 1.5 k [2 point]
STEP5: find VC(t) [0.5 point]
VC(t)=Vc() + (Vc(0)-Vc())e (-t/Tc)
= 27 + 9 e(-6.667t) V
1 point for the plot. You lose 0.5
point for not labeling the axes
properly and/or for not indicating
the initial and final values of Vc(t)
on the graph.
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