NAROK UNIVERSITY COLLEGE
(A CONSTITUENT COLLEGE OF MOI UNIVERSITY)
UNIVERSITY EXAMINATIONS 2011/2012
SECOND YEAR FIRST SEMESTER EXAMINATION
SCHOOL OF SCIENCE
BACHELOR OF COMPUTER SCIENCE
COURSE CODE: COM 215
COURSE TITLE: ELECTRIC CIRCUITS
DATE:
DECEMBER , 2011
TIME: 3 HOURS
INSTRUCTIONS
 Answer Question ONE and any other TWO.
 Use of sketch diagrams where necessary and brief illustrations are encouraged.
 Read the instructions on the answer booklet keenly and adhere to them.
This paper consists of 7_ printed pages.
Page 1 of 7
QUESTION ONE: [24 marks]
a) Define the following terms used in electric circuits
(i) Parameter
[1]
(ii) Electric Network
[1]
(iii) Bilateral circuit
[1]
 Parameter: Are the various elements of an electric circuit like resistance, inductance and
capacitance
 Electric Network – is a combination of various electric elements connected in any manner
 Bilateral circuit – is the one whose properties or characteristics are the4 same in either
direction e.g transmission
b) State the following theorems:
(i)
Norton’s theorem states
(ii)
Thevenin’s theorem
[2]
[2]
Norton’s theorem states:
 The current that flows in any branch of a network is the same as that which would fl ow in
the branch if it were connected across a source of electrical energy, the short-circuit current
of which is equal to the current that would fl ow in a short-circuit across the branch, and the
internal resistance of which is equal to the resistance which appears across the opencircuited branch terminals.
Thevenin’s theorem
 ‘The current which flows in any branch of a network is the same as that which would flow
in the branch if it were connected across a source of electrical energy, the e.m.f. of which is
equal to the potential difference which would appear across the branch if it were opencircuited, and the internal impedance of which is equal to the impedance which appears
across the open-circuited branch terminals when all sources are replaced by their internal
impedances.’
c) Use Kirchhoff’s’ laws to determine the i, i1, and i2 currents shown in fig.1
[5]
Figure 1
Kirchoff’s Rules:
V = 0
i
in any loop and
i
in
=  i out at any junction
i
Applyrule 1: 12  4i1  3i  0 and again  2i2  5  4i1  0; Apply rule 2 : i  i1  i2
Page 2 of 7
39  26i1  0
12  4i1  3(i1  i 2 )  0
12  7i1  3i 2  0
and solving the equations gives
5  4i1  2i 2  0
39
 1. 5 A
26
i 2  0. 5 A
i1 
i  2.0 A
d) A 3.33-kΩ resistor is connected to a generator with a maximum voltage of 101 V. Find:
(i) the average and
[2]
(ii) the maximum power delivered to this circuit.
[2]
(i) Vrms  Vmax / 2   101V / 1.414  71.4V
(ii)
2
2
P  Vrms
/ R  (71.4V)2 / (3330)  1.53W and PMax  Vmax
/ R  (101V) 2 / (3330)  3.06W
e) Find the current phasor if a 50 Hz 22030o V ac voltage is applied to
(i) a pure resistive circuit of R=10 
[2]
(ii) a pure inductive circuit of L=0.2 H
[3]
(iii)a pure capacitive circuit of C=10 F
[3]
0
V 22030
i) I R  
 22300 A
R
10
V
220300
220300
ii) XL  j2fL  j2(50)(0.2)  j62.83  and I L 


 3.50  600 A
0
XL
j62.83
62.8390
1
1
V 220300
220300
iii) XC   j
j
  j318.3 ; I C 


 0.6911200 A
6
0
2fC
XC
 j318.3 318.3  90
2(50)(10x10 )
f) 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.
[6]
Let Z1  (2  j5), Z2  (6  j3), Z3  (3  j4)  Z1Z2  6  j33 and Z3Z1  59  j2
Z3 Z2
Z Z
I1  I
 1.21  j5.55 and I 2  I 3 1  4.36  j1.33
Z1Z2  Z3 Z2  Z3 Z1
 Z1Z2
Z Z
and I3  I 2 1  4.43  j4.22
 Z1Z2
ZI
Z2 Z3Z1
 3.01  j0.51  V  100o  3.059.6o and p.f  cos9.6o  0.986(lag)
Z1Z2  Z3Z2  Z3Z1
QUESTION TWO: [20 marks]
a) 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.
Page 3 of 7
- Step 5: Complete the solutions as required.
b) Use Mesh Analysis to find the current i and power through resistor R; PR in fig. 2. [10]
Figure 2
 Steps 1,2,3,4
By  inspection : i1  15; Apply  KVL to the other  2  meshes :
mesh  2; 1(i 2  i1 )  4(i 2  i 3 )  0  i 2  15  4i 2  4i 3  0  5i 2  4i 3  15
mesh  3; 4(i 3  i 2 )  2i 3  10  0  3i 3  2i 2  5
25
5
20
,i 3   i R  i 2  i 3 
,  PR  32.7W
7
7
7
c) Explain the procedure for the application of the Millman’s Theorem
[5]
 Any number of parallel voltage sources can be reduced to one.
 This permits finding the current through or voltage across RL 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.
QUESTION THREE: [20 marks]
a) What is a Wheatstone bridge smoke detector?
[3]
 A high input impedance circuit is employed to sense the condition of a bridge such as a
Wheatstone bridge containing ionization chambers for smoke detection.
b) Explain the construction and working of Wheatstone bridge smoke detector [17]
 i2 
 A high input impedance circuit is employed to sense the condition of a bridge such as a
Wheatstone bridge containing ionization chambers for smoke detection. The circuit comprises two
CMOS inverters connected to operate as a differential signal detector, one inverter connected at its
Page 4 of 7
gate electrodes to one output terminal of the bridge and the other similarly connected to the other
output terminal of the bridge. The system includes a simple null and sensitivity adjustment.
QUESTION FOUR: [20 marks]
a) Explain the following
(i) Capacitive reactance
[2]
(ii) Inductive reactance
[2]
(iii)Impedance
[2]
(iv) Phasor
[2]
 The “resistance” of the capacitor to current in the circuit is known as capacitive reactance
XC = 1/(wC)
 The “resistance” of the inductor to current in the circuit is known as inductive reactance
XL = 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?
[1]
(iii)Calculate the current for a frequency of 2.00 kHz.
[1]
1
s
s
V
(i) XC  1 / (C) 
 0.403  0.403
 0.403  0.403
3
F
C/ V
A
( 2 )(1000Hz)(0.395  10 F)
Irms  Vrms / XC  10.0V / (0.403)  0.0248A  24.8A
(ii)
If frequency is doubled, XC drops by factor of 2; hence current is doubled.
(iii) Irms = 49.6 mA at 2.00 kHz
d) For the circuit in fig. 3
(i) Compute PT and QT for the following circuit.
(ii) Reduce the circuit to its simplest form
Figure 3
Solution :
(i) PT  I 2 R  ( 20)2 (3)  1200 W
Q C1  I 2 XC1  ( 20)2 ( 6)  2400 VAr (cap.)
and
V2
2002

 4000 VAr (cap.) and
XC 2
10
QT = -2400 – 4000 +8000 = 1600 VAR
QC 2 
QL 
Page 5 of 7
V 2 2002

 8000 VAr (ind.)
XL
5
[5]
[2]
(ii) Req  R  3 
QUESTION FIVE:
and
Xeq 
QT 1600

 4  (ind.)
I2
202
[20 marks]
a) Define power factor angle; θ
[3]
 Power factor angle  is the angle between applied voltage and current or the angle between
S and P
b) Show that for the circuit shown on fig.4 , Current lags voltage by 90o
Figure 4
di
d
 L (I m sin t)  LI m cos t  LI m sin(t  90o )  Vm sin(t  90o )
dt
dt
where Vm  LI m
vL  L
Voltage & current phasors 1
Voltage & current waveforms 1
  Current lags voltage by 90o
Page 6 of 7
c) Refer to the circuit of Fig 5, find:
(i) the total impedance, ZT
(ii) the supply current, IT
(iii) the branch currents, I1, I2 and I3.
[6]
[2]
[5]
Figure 5
Solution :
1
j
 0.0500 mS;
Y2 
 1  900 mS ;
3
0
1000
20x10 0
j
Y3 
 1900 mS; and YT  Y1  Y2  Y3  0.0500 mS 
1000
(i) Y1 
ZT 
1
 R  2000 k
YT
(ii) I T  YT E  ( 0.05x103 00 )(500 )  0.2500 mA
(iii) I1  Y1E  I T  0.2500 mA also
I 2  Y2 E  (0.001  900 )(500 )  5  900 mA
and I 3  Y3 E  (0.001900 )(500 )  5900 mA
End and Good Luck.
Page 7 of 7