ECE 2215 ELECTRIC CIRCUIT ANALYSIS I
CAT 1 [35 MARKS]
TIME: 1 HOUR
Instructions: Attempt ALL questions
1. For the series-parallel network shown in Fig, 1, find
a) Total circuit resistance
b) The supply current
c) The current flowing through each resistor
d) The p.d. across each resistor
e) The total power dissipated in the circuit
2
R2
8
2.4 
R4 I
R1
I1
3.6 
I
R3
[8 marks]
R5
I2
5
100V
Fig. 1
Solution:
The equivalent resistance for resistor connected in parallel is
10  5 10
R p = ( R1 + R2 ) //R3 = ( 8 + 2 ) //5 = 10 //5 =
=

10 + 5
3
Total circuit resistance is
10
28
RT = R 4 + R p + R5 = 2.4 +
+ 3.6 =
 or 9.33 
3
3
a) The supply current
V
100 3  100
I =
=
=
= 10.71A
28
RT
28
3
b) The current flowing through each resistor
(i) Current through R5 = I = 10.71A
(ii) Current through R4 = I = 10.71A
[1 mark]
[1 mark]
[½ mark]
[½ mark]
 8+2 
 8+2 
(iii) Current through R3 , I 2 = 
I = 
10.71A=7.14 A
8+2+5
8+2+5
[½ mark]
5
5




(iv) Current through R1 & R2 , I1 = 
I = 
10.71A=3.57 A
8+2+5
8+2+5
[½ mark]
Page 1 of 6
c) The p.d. across each resistor
(i)
Voltage across R5 ,VR5 = IR5 = 10.71 3.6 = 38.56 V
[½ mark]
(ii)
Voltage across R4 ,VR4 = IR4 = 10.71 2.4 = 25.7 V
[½ mark]
(iii)
Voltage across R3 ,VR3 = I 2R3 = 5  7.14 = 35.7 V
[½ mark]
(iv)
Voltage across R2 ,VR2 = I1R2 = 2  3.57 = 7.14 V
[½ mark]
(v)
Voltage across R1,VR1 = I1R1 = 8  3.57 = 28.56 V
[½ mark]
d) The total power dissipated in the circuit
PT = I 2R4 + I 2R4 + I 22R3 + I12R2 + I12R1
= (10.71) 3.6 + (10.71) 2.4 + ( 7.14) 5 + (3.57 ) 2 + (3.57 ) 8
2
2
2
2
2
= 1070.57 W
[1½ marks]
2. Using mesh method in circuit shown below to find
a)
b)
c)
d)
The currents I1 , I 2 and I 3
the current given by the battery
the voltage across resistor at branch RS and
current in branch PR
8k 
P
0.5 k 
I3
10 k 
20 k 
120 V
[9 marks]
Q
10 k 
I2
I1
S
15 k 
R
Solution:
a) Let the currents I 1 , I 2 and I 3 be the mesh currents in mA.
Applying KVL to loop 1; we get
120 − 0.5I1 − 20 ( I1 − I 2 ) = 0
20.5I1 − 20I 2 = 120
(i)
[1 mark]
Applying KVL to loop 2; we get
−20 ( I 2 − I1 ) − 10 ( I 2 − I 3 ) − 15I 2 = 0
20I1 − 45I 2 + 10I 3 = 0
(ii)
Applying KVL to loop 3; we get
−8I 3 − 10I 3 − 10 ( I 3 − I 2 ) = 0
Page 2 of 6
[1 mark]
10I 2 − 28I 3 = 0
(iii)
[1 mark]
Solving for I 1 , I 2 and I 3 using equs. (i), (ii) and (iii), we get
6960
mA or I1 = 11.07 mA
629
3360
I2 =
mA or I 2 = 5.34 mA
629
1200
I3 =
mA or I 3 = 1.907 mA
629
b) the current given by the battery
I1 =
[1 mark]
[1 mark]
[1 mark]
I1 = 11.07 mA
c) the voltage across RS and
[½ mark]
I RS = I 2 = 5.34 mA
VRS = (15  10 ) I RS = (15  10
3
3
)(5.34 10 ) = 80.1V
−3
[½ mark]
[1 mark]
d) current in branch PR
I PR = I 2 − I 3 = (5.34 − 1.907 ) mA=3.433 mA
[1 mark]
3. State the Kirchhoff’s laws and explain the steps used in solving a circuit
using Kirchhoff’s laws.
[5 marks]
a) KCL states that in any network of conductors, the algebraic sum of currents
meeting at a point (or a junction) is zero i.e. the total current leaving a
junction is equal to the total current entering that junction.
[1 mark]
b) Kirchhoff’s Voltage Law states that the algebraic sum of all IR drops and
emfs in any closed loop (or mesh) of a network is zero i.e.
 IR +  e.m. f = 0
▪
[1 mark]
To solve a circuit using Kirchhoff’s laws, the following steps are used:-
1. According to KCL, mark the direction of flow of current in various branches
of the circuit.
[1 mark]
2. Select as many numbers of closed circuits as the number of unknown
quantities.
[1 mark]
3. For the selected closed circuits, apply KVL and write down the equations.
[½ mark]
4. Calculate the unknown values, solving the equations written in step (iii)
[½ mark]
Page 3 of 6
4. In the circuit of Fig. 2, find
a) The currents I 1 and I 2
b) The voltage across 9  .
c) Power dissipated in the 15  .
2.5Ω
A
5Ω
B
I1
[7 marks]
I2
I1 + I 2
10V
0.5Ω
C
25V
1Ω
9Ω
15Ω
D
Fig.2
Solution:
a) Apply KVL to loop ABDA,
10 − 0.5I1 − 2.5I1 − 9(I1 + I 2 ) = 0 or 12I1 + 9I 2 = 10
Apply KVL to loop BCDB,
5I 2 − 25 + I 2 + 15I 2 + 9(I1 + I 2 ) = 0 or 9I1 + 30I 2 = 25
(i)
[1 mark]
(ii)
[1 mark]
Solving for I 1 and I 2 using equations (i) and (ii), we get
25
A or 0.269 A
93
70
I2 =
A or 0.753 A
93
b) The voltage across 9  .
I1 =
[1 mark]
[1 mark]
 25 70 
V9 = 9 ( I1 + I 2 ) = 9 
+
 =9.19 V
 93 93 
c) Power dissipated in the 15  .
[1½ marks]
2
P15 
 70 
= 15I =15 
 =8.5 W
 93 
2
2
[1½ marks]
5. Use nodal analysis to calculate the current flowing in each branch of the
network shown in below.
[4 marks]
Page 4 of 6
I1
I2
1
I3
25
5
15
80V
30 V
2
Solution:
From the figure above, taking node 2 as the reference node, Applying KCL to
node 1, have: I 3 = I1 + I 2
80 − V1 30 − V1 V1
+
=
or
25
5
15
[½ mark]
80 30
1 1 1 
+ = V1  + + 
25 5
 5 15 25 
1 1 1 
9.2 = V1  + + 
 5 15 25 
23
9.2 = V1
75
V1 = 9.2 
75
= 30V
23
[½ mark]
I1 =
80 − 30
= 2A
25
[1 mark]
I2 =
30 − 30
= 0A
5
[1 mark]
I3 =
30
= 2A
15
[1 mark]
6. Define the following terms as used in circuit analysis:
a) Circuit – it is a network which contains one or more closed paths around
which current can flow.
[1 mark]
Page 5 of 6
b) Branch - is a portion of a network which contains one or more circuit
elements connected in series.
[1 mark]
Page 6 of 6