(A CONSTITUENT COLLEGE OF MOI UNIVERSITY)

ONE
TWO


This paper consists of 7_ printed pages.
Page 1 of 8

QUESTION ONE: a) Define the following terms used in electric circuits
(i) Passive Element [1]
[30]
(ii) Electric Network [1]
(iii) Active Element [1]

Passive Element: The element which receives energy (or absorbs energy) and then either converts it into heat (R) or stored it in an electric (C) or magnetic (L ) field is called passive element.

Electric Network – is a combination of various electric elements connected in any manner.

Active Element: The elements that supply energy to the circuit is called active element. Examples of active elements include voltage and current sources, generators, and electronic devices that require power supplies. A transistor is an active circuit element, meaning that it can amplify power of a signal b) A total current of 10 A flow through the parallel combination of three impedance: (2-5j)Ω,
(6 + 3j) Ω, and (3 + 4j) Ω. Calculate the current flowing through each branch. Find also the p.f of the combination. [5]
Let Z
1

(2
 j5), Z
2

(6
 j3), Z
3
(3 j4)

Z Z 6 j33 and Z Z

59
 j2
I
1

I
Z Z

Z Z

Z Z

1.21
 j5.55 and I
2

I 

4.36
 j1.33
and I
3

I 

4.43
 j4.22
Z

I
Z Z

Z Z

Z Z

3.01
 j0.51
     o and p.f
 cos 9.6
o 
0.986(lag) c) Explain the steps involved in applying the Mesh Analysis in solving electric circuits [5]
Step 1: Determine & give current values to each mesh.
Step 2: Where there are current sources; determine some mesh currents by inspection.
Step 3: Use KVL to write the other mesh equations & solve them simultaneously
Step 4: Determine the resultant current through each resistor.
Step 5: Complete the solutions as required. d)
Use Kirchhoff’s’ laws to determine the i, i
1
, and i
2 currents shown in fig.1 [5]
Page 2 of 8

Figure 1
Kirchoff’s Rules:
 i
V = 0 in any loop and
 i = in
 i out at any junction
Applyrule
12

4 i
1

3
1: 12 4
(i
1
 i )
12

7 i
1

3 i
2

0
5 4 i
1

2 i
2

0
2
 i
1
3 i 0 and again  2 i
2
  i
1
 0; Apply rule
39

26 i
1

0
2 :   i
2
0 and solving the equations gives i i
2
1

39

26
 i

2 0 e)
Explain the procedure for the application of the Millman’s Theorem [4]

Any number of parallel voltage sources can be reduced to one.

This permits finding the current through or voltage across R
L
without having to apply a method such as mesh analysis, nodal analysis, and superposition and so on. i.
Convert all voltage sources to current sources. ii.
Combine parallel current sources. iii.
Convert the resulting current source to a voltage source and the desired singlesource network is obtained. f) Find the current phasor if a 50 Hz 220

30 o V ac voltage is applied to
(i) a pure resistive circuit of R=10

(ii) a pure inductive circuit of L=0.2 H
(iii)a pure capacitive circuit of C=10

F i) I
R

V

R
 0
10
  0
A ii) X
L
  fL j 2 ( 50 0 2
 j 62 83
 and I
L

V
X
L
 j

.
0

.


0
0 iii) X
C
 
1 j
2
 fC
  j
1
 
6
)
  j 318 3

; I
C

V
X
C

 0
 j 318 3


 0
.
 
90
0
[2]
[3]
3 50
 
60
0

[3]
.
 0
A
A
Page 3 of 8

QUESTION TWO: [20] a) Summarize the steps to analyze a circuit by node voltage method. [10]

Step-1: Identify all nodes in the circuit. Select one node as the reference node (assign as ground potential or zero potential) and label the remaining nodes as unknown node voltages with respect to the reference node.

Step-2: Assign branch currents in each branch. (The choice of direction is arbitrary).

Step-3: Express the branch currents in terms of node assigned voltages .

Step-4: Write the standard form of node equations by inspecting the circuit. (No of node equations = No of nodes (N) – 1).

Step-5: Solve a set of simultaneous algebraic equation for node voltages and ultimately the branch currents. b) Use Nodal analysis to obtain i, P
R
in fig. 2. [10]
Figure 2
 Steps 1,2,3,4
Apply-KCL-at-node-of-v: but
 v
1


;by
 inspection; v

v
1 v v

 v

v v


v

v
v

v
i v
R



P
R
 i R

Page 4 of 8

QUESTION THREE: [20 marks] a) Show that for the circuit shown on fig.3 , Current lags voltage by 90 o
Figure 3
[4] v
L

L di dt where V m
 d
L (I sin t) LI cos t LI sin( t m m m dt
       
 
LI m
90 o
)
 m
 
90 o
)
Voltage & current phasors 1
Voltage & current waveforms 1
 
Current lags voltage by 90 o b) Refer to the circuit of Fig 4, find:
(i) the total impedance, Z
T
(ii) the supply current, I
T
(iii) the branch currents, I
1
, I
2
and I
3
.
[7]
[3]
[6]
Figure4
Page 5 of 8

Solution :
(i) Y
1

1 x
3 
0
0

.
 0 mS; Y
2

 j
1000
1 90 0 mS ;
Y
3
 j
1000
0 mS; and Y
T

Y
(ii) I
T

Y E

( .
x

3 
0
0
)(
 0
)

.
1


Y
2
0

Y mA
3
  0 mS

Z
T

1
Y
T

R

20 0
0 k

(iii) I
1

Y E
1

I
T
  0 and I
3

Y E
3

( .
mA also I
2

Y E
2

( .
 
90
0
)(
 0
)
  
90
0
 0
)(
 0
)
  0 mA mA
QUESTION FOUR: [20 marks] a) Explain the superposition principle [6]
 The response of a linear circuit, acted upon by several independent sources, is the algebraic sum of the responses when each source acts alone”.
 The Principle is useful either in analyses where it is difficult to use the above methods easily or when the individual effects of sources are needed. b) Use the Superposition Principle to determine the current in R in figure 5. [14]
Figure 5

Let @ source act alone. Nodal Analysis:
39 A current
 
: v
1

26
3 v
1 v
4 2
1 0
: v
1
8 i
1
2 v
2
 v
2
 v
3 4 2
2

39 0 v
2
36
13 i
A current i
1 i
2 i
3
 
: v
3
 v
3
 v
3 4 2
3

13 0 v
3

12 i
3
2 9 3 4 4 A upwards .
i
2
3
Page 6 of 8
9

QUESTION FIVE: [20 marks] a) Explain the following
(i) Capacitive reactance
(ii) Inductive reactance
(iii)Impedance
[2]
[2]
[2]
(iv) Phasor [2]
 The “resistance” of the capacitor to current in the circuit is known as capacitive reactance
X
C
= 1/(wC)
 The “resistance” of the inductor to current in the circuit is known as inductive reactance
X
L
= wL

Impedance is the total opposition to an electrical circuit from the combined resistance, and reactance.

A phasor is a complex number whose magnitude is the magnitude of a corresponding sinusoid, and whose phase is the phase of that corresponding sinusoid. b) An rms voltage of 10.0 V with a frequency of 1.00 kHz is applied to a 0.395-mF capacitor.
(i) What is the rms current in this circuit? [3]
(ii) By what factor does the current change if the frequency of the voltage is doubled?
(iii)Calculate the current for a frequency of 2.00 kHz.
[1]
[1]
(i) X
C

1 / ( C)


V / X rms C
1
( 2

)( 1000
 
3
F)

.
s
.
F


10 0 0 403
 
0 0248 A

. A s
C / V

I rms
(ii) If frequency is doubled, X
C
drops by factor of 2; hence current is doubled.
(iii) I rms
= 49.6 mA at 2.00 kHz c) For the circuit in fig.6
(i) Compute P
T
and Q
T
for the following circuit.
(ii) Reduce the circuit to its simplest form
Figure 6
V
A
 
[5]
[2]
Solution :
(i) P
T
 2
I R

( 20
2
3

1200 W and
V
2
Q
C 2

X
C 2

200
2
10

4000 VAr (cap.) and
Q
T
= -2400 – 4000 +8000 = 1600 VAR
Q
C 1
 2
I X

( ) ( )
C 1
20
2
6

2400 VAr (cap.)
Q
L

V
X
L
2

200
2
5

8000 VAr (ind.)
Page 7 of 8

(ii) R eq

R 3 and X eq

Q
T
I 2

1600
20 2
End and Good Luck.
(ind.)
Page 8 of 8