[EEE 2019 TEST 1 SOLUTIONS 2022/2023 Academic Year]
[Dept. of EEE, School of Engineering, University of Zambia]
EEE 2019 - PRINCIPLES OF ELECTRICAL AND ELECTRONIC ENGINEERING
TEST 1 [SOLUTIONS] – TERM 1 2022/2023 ACADEMIC YEAR – APRIL 25, 2023
QUESTION 1
(a) The current entering the positive terminal of a device is i(t )  3e 2t A and the
voltage across the device is v(t )  5
di(t )
V.
dt
t  0 and t  2s , given
(i) Determine the charge delivered to the device between
[4 Marks]
that q(0)  0 .
(ii) Find the power absorbed.
[3 Marks]
(iii) Determine the energy absorbed in 3s .
[4 Marks]
[SOLUTION] [Q1(a)]
(i) Given that, i(t )  3e 2t A it follows that,
t

e 2t 

 q(t )  i(t )dt  q(0) ; i.e., q(t )  3 e dt  0  3 
 ;
0
0

2



0

t
t
3
2
 Thus, q(t )   (e
2t
2t
 e 0 ) ; i.e., q(t )  3 (1  e 2t ) C ,
[2 Marks]
2
3
 At t  2s , we obtain, q(2)  (1  e 4 ) C  1.4725 C .
2
(ii) Given that, v(t )  5
 v(t )  5
[2 Marks]
di(t )
V , we obtain
dt
d
d
(3e 2t )  5(6e 2t ) V ; i.e., v(t )  5 (3e 2t )  30e 2t V ;
dt
dt
 It follows that, p(t )  v(t )i(t )  (30e 2t V)(3e 2t A) ; i.e., p(t )  90e 4t W ;
 At t  2s , we obtain, p(2)  90e 8 W  0.0302 W ;
[2 Marks]
[1 Mark ]
(iii) Energy in terms of power is of the form,
 w(t ) 

t
0
p(t )dt  w(0) ;
t

e 4t 

 Assume, w(0)  0 , it follows that, w(t )   p(t )dt  90 e dt  90 
 ;
0
0

 4 
0
45
[2 Marks]
 Thus, w(t )   (1  e 4t ) J ;
2
[2 Marks]
45
 At t  3s , we have, w(3)   (1  e 12 ) J  22.4999 J ;
2
t
t
4t
(b) For the circuit shown in Figure Q1(b), find:
(i) The equivalent resistance Req .
[8 Marks]
(ii) The equivalent conductance Geq .
[3 Marks]
[Prepared by Dr. C. S. Lubobya & Jerry Muwamba]
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[EEE 2019 TEST 1 SOLUTIONS 2022/2023 Academic Year]
[Dept. of EEE, School of Engineering, University of Zambia]
(iii) The value of the current i0 .
[3 Marks]
60 
12 
i0
5
6
80 
15 
40V
20 
Req
Figure Q1(b)
[Total 25 Marks]
[SOLUTION] [Q1(b)]
(i) Consider the given circuit as depicted in FIG. 1,
60 
12 
i0
5
60 
i0
6
5
4
80 
15 
40V
Req
20 
15 
40V
FIG. 1
Req
16
FIG. 2
[2 Marks]
 It follows that FIG. 2 is obtained from the resistor combinations of the form,
(20)(80)
(12)(6)
 16  ;
 4  ; and R  20 80 
100
18
 Vividly, from FIG. 2 we obtain,
 R  12 6 
[2 Marks]
[2 Marks]
1
60
60
15


  7.5  ;
; i.e., Req 
1
2
1
1  [4  3  1] 8
[2 Marks]
 
 
15
20
60


(ii) Therefore, the equivalent conductance is of the form,
 Req  15 (4  16) 60 
 Geq 
1
2
 S  0.1333 S ;
Req 15
[3 Marks]
(iii) It follows from FIG. 2 that,
 i0 
2
40
16
 40   ; i.e., i0 
A  3.2A ;
25 

5
15 

5  
2

[3 Marks]
[Total 25 Marks]
QUESTION 2
(a) In the circuit shown in Figure Q2(a), determine:
[Prepared by Dr. C. S. Lubobya & Jerry Muwamba]
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[EEE 2019 TEST 1 SOLUTIONS 2022/2023 Academic Year]
[Dept. of EEE, School of Engineering, University of Zambia]
(i)
The equivalent resistance Req .
[6 Marks]
(ii)
The voltage vx across the 1 ohm resistor.
[4 Marks]
(iii)
The power absorbed by the 2 ohm resistor.
[3 Marks]
1
1.2 
vx
4
2
6A
8
3
12 
6
Figure Q2(a)
[SOLUTION] [Q2(a)]
(i) Consider the given circuit as depicted in FIG. 1,
1
vx
6A
2
1.2 
a
1
b
vx
4
8
c
3
d
a
12 
6A
2
b
4
4.8 
c
6
d
1.2 
2
d
FIG. 1
d
d
FIG. 2
[2 Marks]
 It follows that FIG. 2 is obtained from the resistor combinations of the form,
(8)(12) 24
(3)(6)

  4.8  ;
 2  ; and R  8 12 
20
9
5
 Vividly, from FIG. 2 we obtain,
 R  3 6 
[2 Marks]
4
(2)(4)
; i.e., Req    1.3333  ;
3
6
(ii) Therefore, the voltage vx across the 1 ohm resistor is of the form,
[2 Marks]
 Req  2 (1  6 6)  2 4 
 2 
 vx  (1)(i1 ) ; where i1  
[4 Marks]
 (6)  2 A ; i.e., vx  (1)(2)  2 V ;
2  4 
(iii) It follows from FIG. 2 that the power absorbed by the 2 ohm resistor is,
 p2   v2 i2   Ri22  (2)(6  2)2  (2)(16) ; i.e., p2  32W ;
[3 Marks]
(b) Using delta to wye transformation, obtain the equivalent resistance RAB between
terminals A and B.
[Prepared by Dr. C. S. Lubobya & Jerry Muwamba]
[12 Marks]
Page 3 of 9
[EEE 2019 TEST 1 SOLUTIONS 2022/2023 Academic Year]
[Dept. of EEE, School of Engineering, University of Zambia]
A
30 
RAB
30 
30 
30 
30 
30 
20 
B
Figure Q2(b)
[Total 25 Marks]
[SOLUTION] [Q2(b)]
 Firstly, we convert the two balanced deltas to wye as illustrated in FIG. 1, which
yields the circuit of the form of FIG. 2.
A
A
R  30 
R
RAB
RY  10 
RAB
RR
R
RY
RY
20
B
B
RY
R
FIG. 1
RY
RY
FIG. 2
20
[3 Marks]
 For the balanced deltas in FIG. 1, we obtain resistances for the wyes as,
 RY 
R
3

30
 10 ; i.e., RY  10  ;
3
[3 Marks]
 From FIG. 2, we obtain,
(20)(40) 
 Rab  RY  (2RY ) (2RY  20)  RY ; i.e., Rab  20  20 40  20  
;
 60 
[3 Marks]
100
40 100
[3 Marks]
  33.333  ;

;  R 
 Thus, R  20 
ab
3
ab
3
3
[Total 25 Marks]
QUESTION 3
(a) Using nodal analysis in the circuit of Figure Q3(a), determine
(i)
The voltage V0 .
[10 Marks]
(ii)
The current I x .
[3 Marks]
1
10 
Ix
30V
2
4I x
5
V0
Figure Q3(a)
[SOLUTION] [Q3(a)]
(i) Step 1, we assign node voltages to each nonreference node as depicted in FIG. 1.
[Prepared by Dr. C. S. Lubobya & Jerry Muwamba]
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[EEE 2019 TEST 1 SOLUTIONS 2022/2023 Academic Year]
V1
10 
[Dept. of EEE, School of Engineering, University of Zambia]
1
V2
V0
Ix
2
30V
4I x
5
V0
datum
FIG. 1
[2 Marks]
 At node 1, by inspection we obtain
[2 Marks]
 V1  30V ;
 At node 2, applying KCL yields,

 30  V2 
 
V2 
 
V2  V0 

 
 
  0 ; i.e., 30  V2  5V2  10V2  10V0  0 ;
 10 
 
2
 
 1 


 Thus, 10V0  16V2  30 ; i.e., 5V0  8V2  15 ;
(1)
[2 Marks]
 At node 0, applying KCL gives,


V  V0 

V2 
 
V0 

  2
  4       0 ; i.e., 5V2  5V0  10V2  V0  0 ;


 1 

2
 
5

 Thus, 6V0  15V2  0 ; i.e., 2V0  5V2  0 ;
(2)
[2 Marks]
 Substituting eq(2) into eq(1) yields,
 25  16 
2 
5
 5V0  8  V0   15 ; i.e., 
V0  15 ; V0  15   ;
5

5 

9
25
V  8.3333V ;
 Therefore, V0  
3
(ii) It follows from (i) above that,
[2 Marks]
V
2
 2   25  10
5
 V2   V0      
V ; i.e., I x  2  A  1.6667 A ;
2
3
3
5
 5  3 
[3 Marks]
(b) Consider the circuit shown in Figure Q3(b). Determine the current, voltage, and
power associated with the 20 k .
5mA
10k
[12 Marks]
V0
0.01V0
5k
20k
Figure Q3(b)
[Total 25 Marks]
[SOLUTION] [Q3(b)]
 We may use nodal analysis on FIG. 1 as follows,
[Prepared by Dr. C. S. Lubobya & Jerry Muwamba]
Page 5 of 9
[EEE 2019 TEST 1 SOLUTIONS 2022/2023 Academic Year]
[Dept. of EEE, School of Engineering, University of Zambia]
V1
V0
5mA
10k
V0
5k
0.01V0
20k
FIG. 1
 At node 0, applying KCL yields,
 V 
0
5mA



  0 ; i.e., V0  (10k)(5mA)  50V ;
10k



[3 Marks]
 At node 1, applying KCL yields,
V   V 
1
1

  0 ; i.e., (20k)0.01V0  4V1  V1  0 ;  5V1  (20k)0.01V0 ;
5k
20k
  

[3 Marks]
 Thus, V1  (4k)0.01V0  (4k)(0.01)(50) ;  V1  V20k  2kV ;
 0.01V0  
 It follows that, p20k 
2
v20k

R

(2k)2
20k

4k
20
; i.e., p20k 
1
kW  0.2kW ; [4 Marks]
5
[Total 25 Marks]
QUESTION 4
(a) Find V0 and the power absorbed by each element in the circuit of Figure Q4(a).
[13 Marks]
28V
I 0  2A
12V
6A
30V
p1
p5
1A
p2
V0
3A
p3
28V
p4
p6
5I 0
3A
6A
Figure Q4(a)
[SOLUTION] [Q4(a)]
 Applying KVL around the loop yields,
[1 Mark ]
 V0  30  12  18 ; i.e., V0  18 V ;
 For p1 we obtain,
 p1  vi  (30)(6)  180 ; i.e., p1  180W;
supplied ;
[2 Marks]
 For p2 we obtain,
 p2  vi  (12)(6)  72 ; i.e., p2  72W;
[Prepared by Dr. C. S. Lubobya & Jerry Muwamba]
absorbed ;
[2 Marks]
Page 6 of 9
[EEE 2019 TEST 1 SOLUTIONS 2022/2023 Academic Year]
[Dept. of EEE, School of Engineering, University of Zambia]
 For p3 we obtain,
 p3  vi  (18)(3)  54 ; i.e., p3  54 W;
absorbed ;
[2 Marks]
absorbed ;
[2 Marks]
absorbed ;
[2 Marks]
 For p4 we obtain,
 p4  vi  (28)(1)  28 ; i.e., p4  28 W;
 For p5 we obtain,
 p5  vi  (28)(2)  56 ; i.e., p5  56W;
 For p6 we obtain,
 p6  vi  (5I 0 )(3)  (5)(2)(3)  30 ; i.e., p6  30W;

p
supplied
supplied ;
[2 Marks]
  pabsorbed  0 ; LHS, (180  30)  (72  54  28  56)  0 ; Q.E.D,
(b) For the circuit shown in Figure Q4(b), using mesh analysis determine
(i) The value of the currents i1 and i2 using Cramer’s rule.
[10 Marks]
(ii) The value of the voltage v 0 .
[2 Marks]
3
6
v0
12V
8
i1
i2
2v 0
Figure Q4(b)
[Total 25 Marks]
[SOLUTION] [Q4(b)]
(i) With respect to the circuit in FIG. 1 under analysis we have,
3
6
v0
12V
8
i1
i2
2v 0
FIG. 1
 Mesh 1, applying KVL yields,
 12  3i1  8(i1  i2 )  0 ; i.e., 11i1  8i2  12 ;
 Mesh 2, applying KVL yields,
(1)
[2 Marks]
 8(i1  i2 )  6i2  2v0  0 ; where v0  3i1 ;
 Thus, 8(i1  i2 )  6i2  2(3i1 )  0 ; i.e., 2i1  14i2  0 ;  i1  7i2  0 ;
 Using Cramer’s rule we put the simultaneous equations in matrix form,
11 8  i1  12
11 8
 
 77  (8)  69 ;
      ; i.e.,  
1

7
i
0
1

7

  2   
[Prepared by Dr. C. S. Lubobya & Jerry Muwamba]
(2)
[2 Marks]
(3)
Page 7 of 9
[EEE 2019 TEST 1 SOLUTIONS 2022/2023 Academic Year]
 1 
[Dept. of EEE, School of Engineering, University of Zambia]
12 8
11 12
 12 ;
 84 ; and 2 
0 7
1 0
(4)
[2 Marks]
 It follows that,
1
28
84 28

; i.e., i1 
A  1.2174A ;

69 23
23

4
12
4
A  0.1739A ;

 i2  2 
; i.e., i2 

69 23
23
 i1 

[2 Marks]
[2 Marks]
(ii) Thus, value of the voltage v 0 is the form,
 28 
84
V  3.6522 V ;
 v0  3i1  3   ;  v 0 
23
 23 
[Prepared by Dr. C. S. Lubobya & Jerry Muwamba]
[2 Marks]
[Total 25 Marks]
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[EEE 2019 TEST 1 SOLUTIONS 2022/2023 Academic Year]
[Dept. of EEE, School of Engineering, University of Zambia]
END OF EEE 2019 TEST I [SOLUTIONS]
[Prepared by Dr. C. S. Lubobya & Jerry Muwamba]
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