ELECTRICAL NETWORKING PROBLEMS SOLVED BY SYSTEM OF LINEAR EQUATIONS, ELECTRICAL CIRCUITS OR NETWORK THEORY

1. APPLICATION
OF SYSTEM OF
LINEAR
ALGEBRA
ELECTRICAL NETWORK PROBLEMS
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35. An electrical system may be in the form of a circuit or a network. The essential
componentsofsuchasystemareresistors,capacitors(oracondensers),aninductorsand
batteriesoranyothersourceofvoltageincludinganelectromotiveforce(e.m.f.).
Aresistoruseselectricalenergyandconvertsitintoheat.Anelectriclightbulb,electric
tubeandelectricheaterareexamplesofresistors.Acapacitorstoreselectricalenergyand
aninductoropposeachangeinelectricalcurrent.Theelectromotiveforce(suppliedbya
batteryorgenerator)isthecauseofflowofcurrent.
Symbols,GraphicalRepresentationandUnits
Electrical Networks
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36. C R L V
(a) (b) (c) (d)
Battery
In a circuit: (e)
(i) a capacitor is denoted by C, its capacitance is measured in farads (F) and its graphical
representation is shown in figure (a).
(ii) A resistor is denoted by R, its resistance is measured in ohms ( ) and its graphical
representation is displayed in figure (b).
(iii) An inductor is denoted by L, its conductance is measured in henries (H) and its
graphical representation is displayed in figure (c).
(iv) The electromotive force is measured in volts (V). The e.m.f. of a battery is fixed and
may be denoted by E. The e.m.f. of a generator varies periodically and may be denoted
by E0 sin wt where w is the angular frequency. Graphical representations of a
battery and generator are displayed in figure d and e.
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37. Electrical Quantities
The common electrical quantities are charge (quantity of electricity) denoted by q; current
i.e. the amount of charge flowing per second through the cross section of a wire, denoted
by ;dtdqI = where the time t is measured in seconds. The charge is measured in
coulombs and the current in amperes.
The flow of current in a circuit is governed by Ohm’s law and Kirchoff’s laws which can
be stated as follows:
Ohm’s Law: The voltage drop across a resistor is the product of the current and the
resistance.
Kirchoff’s First Law (Current Rule): The sum of the currents flowing into a node is
equal to the sum of the currents flowing out.
Kirchoff’s Second Law (Voltage Rule): The algebraic sum of the voltage drops around
a closed circuit is equal to the total voltage in the circuit.
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38. REMARK: When applying the current rule, select a direction (clockwise or
counterclockwise)aroundtheclosedcircuitandthenconsiderallvoltagesandcurrents
positiveornegativeaccordingtoastheflowisinthesameoroppositedirection.Thisis
whythetermalgebraicsumisusedinthecurrentrule.
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39. Find currents in each branch of the circuit
shown in the following figure:
6
20V
2
10v
4Ω
1I
𝐼1
𝑰 𝟐
𝐼3
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40. Using Kirchoff’s first law (junction rule) at nodes A and B, we have
0IIIIII 321132 =++−=+ ( )i
and by Kirchoff’s second law (current rule) for the circuits ABCD and ABEF, we have
10I2I4 32 =+− ( )ii
20I6I4 12 =+ ( )iii
Equations (i), (ii) and (iii) can be written in the standard form as
20I4I6
10I2I4
0III
21
32
321
=+
=+−
=−−
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41. The corresponding augmented matrix can be
reduced as follows:
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46. Determine the currents ,I,I 21 and 3I for the electrical network shown in the following
figure.
20 5V
1I
2I 10
3I
10 10V
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47. ApplyingKirchoff’ssecondlawtotheloops,weobtaintheequations:
Applying Kirchoff’s first law, we have += 321 III Thus the above equations can be
written in the standard form as:
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51. : =−== 7.0I,3.0I,4.0I 321
Since I2 is negative, the current flow is from C to B rather than from B to C, as tentatively
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52. Find currents in the circuit shown in the following figure:
C
1I 5
5V
10
2I B
A
5
10V
3I
D
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53. Solution: Kirchoff’s first law applied at junction A gives
0IIIIII 321312 =−+−+=
Similarly, Kirchoff’s second law applied to circuits ABC and ABD gives
5I5I10 12 =+ and −=−− 10I5I10 32
Thus the above equations can be written in the standard form as:
10I5I10
5I10I5
0III
32
21
321
=+
=+
=+−
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57. : ==−= 5/4I,5/3I,5/1I 322 T
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58. For the electrical networks (i) – (iii), determine the currents in the various branches:
4 2V
1I
3 2I
(i)
3I
4 4V
4 1I 3V
2I
(ii)
3
3I
1 4V
1 1I 5V
2 2I
(iii)
1.H.W
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59. This Photo by Unknown Author is licensed under CC BY-SA
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