Robertson C.R. Fundamental electrical and electronic principles
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D.C. Circuits
Chapter 2
Learning Outcomes
This chapter explains how to apply circuit theory to the solution of simple circuits and
networks by the application of Ohm ’ s law and Kirchhoff’s laws, and the concepts of potential
and current dividers.
This means that on completion of this chapter you should be able to:
1 Calculate current fl ows, potential differences, power and energy dissipations for circuit
components and simple circuits, by applying Ohm ’ s law.
2 Carry out the above calculations for more complex networks using Kirchhoff’s Laws.
3 Calculate circuit p.d.s using the potential divider technique, and branch currents using the
current divider technique.
4 Understand the principles and use of a Wheatstone Bridge.
5 Understand the principles and use of a slidewire potentiometer.
31
Resistors cascaded or
connected in series
or
2.1 Resistors in Series
When resistors are connected ‘ end-to-end ’ so that the same current
fl ows through them all they are said to be
cascaded or connected in
series
. Such a circuit is shown in Fig. 2.1 . Note that, for the sake of
simplicity, an ideal source of emf has been used (no internal resistance).
E
V
1
R
1
R
2
R
3
V
2
V
3
I
Fig. 2.1
32
Fundamental Electrical and Electronic Principles
From the previous chapter we know that the current fl owing through
the resistors will result in p.d.s being developed across them. We also
know that the sum of these p.d.s must equal the value of the applied
emf. Thus
VIR VIR VIR
11 2 2 33
volt; volt; and volt
However, the circuit current I depends ultimately on the applied emf E
and the total resistance R offered by the circuit. Hence
EIR
EVV V
volt.
Also, volt
123
and substituting for E, V
1
, V
2
and V
3
in this last equation
we have voltIR IR IR IR
123
and dividing this last equation by the common factor I
RR R R
123
ohm
(2.1)
where R is the total circuit resistance. From this result it may be seen
that when resistors are connected in series the total resistance is found
simply by adding together the resistor values.
Worked Example 2.1
Q For the circuit shown in Fig. 2.2 calculate (a) the circuit resistance, (b) the circuit current, (c) the p.d.
developed across each resistor, and (d) the power dissipated by the complete circuit.
A
E 2 4 V ; R
1
330 ; R
2
1500 ; R
3
470
E
V
1
R
1
R
2
R
3
V
2
V
3
I
330 Ω
1.5 kΩ
470 Ω
24 V
Fig. 2.2
D.C. Circuits
33
(a) R R
1
R
2
R
3
ohm
330 1500 470
R 2300 or 2.3 k Ans
(b)
I
E
R
amp
24
2300
I 10.43 mA Ans
(c) V
1
IR
1
volt
10.43 10
3
330
V
1
3.44 V Ans
V
2
IR
2
volt
10.43 10
3
1500
V
2
15.65 volts Ans
V
3
IR
3
volt
10.43 10
3
470
V
3
4.90 V Ans
Note: The sum of the above p.d.s is 23.99 V instead of 24 V due to the rounding
errors in the calculation. It should also be noted that the value quoted for the
current was 10.43 mA whereas the calculator answer is 10.4347 mA. This latter
value was then stored in the calculator memory and used in the calculations for
part (c), thus reducing the rounding errors to an acceptable minimum.
(d) P EI watt
24 10.43 10
3
P 0.25 W or 250 mW Ans
It should be noted that the power is dissipated by the three resistors in the
circuit. Hence, the circuit power could have been determined by calculating the
power dissipated by each of these and adding these values to give the total. This
is shown below, and serves, as a check for the last answer.
PR
P
P
11
1
11
11 1
I
2
32
2
32
0 43 0 330
35 93
043 0 5
watt
mW
(. )
.
(. ) 000
63 33
0 43 0 470
58
3
32
1
11
11
.
(. )
.
mW
mW
P
total power: watt
so mw
PPP P
P
1 23
250 44.
(Note the worsening e ect of continuous rounding error)
34
Fundamental Electrical and Electronic Principles
Worked Example 2.2
Q Two resistors are connected in series across a battery of emf 12 V. If one of the resistors has a value
of 16 and it dissipates a power of 4 W, then calculate (a) the circuit current, and (b) the value of the
other resistor.
A
Since the only two pieces of data that are directly related to each other concern
the 16 resistor and the power that it dissipates, then this information must
form the starting point for the solution of the problem. Using these data we can
determine either the current through or the p.d. across the 16 resistor (and it
is not important which of these is calculated rst). To illustrate this point both
methods will be demonstrated. The appropriate circuit diagram, which forms an
integral part of the solution, is shown in Fig. 2.3 .
E
V
AB
ABC
16 Ω
V
BC
I
12 V
Fig. 2.3
E 12 V ; R
BC
16 ; P
BC
4 W
(a) I
2
R
BC
P
BC
watt
I
2
P
R
BC
BC
4
6
0.25
1
so I 0.5 A Ans
(b) total resistance,
R
E
I
ohm
12
05.
24 Ω
R
AB
R R
BC
24 16
so R
AB
8 Ans
Alternatively, the problem can be solved thus:
(a)
V
R
P
BC
BC
BC
2
watt
V
BC
2
P
BC
R
BC
4 16
64
D.C. Circuits
35
so V
BC
8 V
I
V
R
BC
BC
amp
8
61
so I 0.5 A Ans
(b) V
AB
E V
BC
volt
12 8
V
AB
4 V
R
V
AB
AB
I
4
05.
so R
AB
8 Ans
2.2 Resistors in Parallel
When resistors are joined ‘ side-by-side ’ so that their corresponding ends
are connected together they are said to be connected in
parallel . Using
this form of connection means that there will be a number of paths
through which the current can fl ow. Such a circuit consisting of three
resistors is shown in Fig. 2.4 , and the circuit may be analysed as follows:
Resistors in Parallel
or or or
I
1
I
I
2
I
3
R
1
R
2
R
3
E
I
Fig. 2.4
36
Fundamental Electrical and Electronic Principles
Since all three resistors are connected directly across the battery
terminals then they all have the same voltage developed across them.
In other words the voltage is the common factor in this arrangement
of resistors. Now, each resistor will allow a certain value of current to
fl ow through it, depending upon its resistance value. Thus
I
E
R
I
E
R
I
E
R
1
1
2
2
3
3
amp; amp; and amp
The total circuit current I is determined by the applied emf and the total
circuit resistance R ,
so ampI
E
R
Also, since all three branch currents originate from the battery, then the
total circuit current must be the sum of the three branch currents
so II I I
123
and substituting the above expression for the currents:
E
R
E
R
E
R
E
R
123
and dividing the above equation by the common factor E :
1111
123
RR R R
siemen
(2.2)
Note:
The above equation does NOT give the total resistance of the
circuit, but does give the total circuit
conductance ( G) which is measured
in Siemens (S). Thus, conductance is the reciprocal of resistance, so
to obtain the circuit resistance you must then take the reciprocal of
the answer obtained from an equation of the form of equation (2.2).
Conductance is a measure of the ‘ willingness ’ of a material or circuit to allow current to
fl ow through it
That is siemen; and ohm
11
R
G
G
R
(2.3)
However, when only two resistors are in parallel the combined
resistance may be obtained directly by using the following equation:
R
RR
RR
12
12
ohm
(2.4)
D.C. Circuits
37
I
1
I
2
I
3
R
1
R
2
R
3
E
I
1.5 kΩ
330 Ω
470 Ω
24 V
Fig. 2.5
In this context, the word
identical means having the
same value of resistance
A
E 2 4 V ; R
1
330 ; R
2
1500 ; R
3
470
(a)
11 1 1
1
RR R R
23
siemen
11
1
1
330 500 470
0.00303 0.000667 0.00213
0.005825 S
so R 171.68 Ans (reciprocal of 0.005825)
(b)
I
1
1
E
R
amp
24
330
I
1
72.73 mA Ans
I
2
2
E
R
amp
24
5001
I
2
16 mA Ans
Worked Example 2.3
Q Considering the circuit of Fig. 2.5 , calculated (a) the total resistance of the circuit, (b) the three branch
current, and (c) the current drawn from the battery.
If there are ‘x ’ identical resistors in parallel the total resistance is
simply
R/x ohms.
38
Fundamental Electrical and Electronic Principles
I
3
3
E
R
amp
24
470
I
3
5 1.06 mA Ans
(c) I I
1
I
2
I
3
amp
72.73 16 5 1.06 mA
so I 139.8 mA Ans
Alternatively, the circuit current could have been determined by using the
values for E and R as follows
I
E
R
amp
24
76811.
I 139.8 mA Ans
Compare this example with worked example 2.1 (the same values
for the resistors and the emf have been used). From this it should be
obvious that when resistors are connected in parallel the total resistance
of the circuit is reduced. This results in a corresponding increase of
current drawn from the source. This is simply because the parallel
arrangement provides more paths for current fl ow.
Worked Example 2.4
Q Two resistors, one of 6 and the other of 3 resistance, are connected in parallel across a source of
emf of 12 V. Determine (a) the e ective resistance of the combination, (b) the current drawn from the
source, and (c) the current through each resistor.
A
The corresponding circuit diagram, suitably labelled is shown in Fig. 2.6.
I
1
I
I
2
12 V
R
2
R
1
E
3 Ω6 Ω
Fig. 2.6
D.C. Circuits
39
E 12 V ; R
1
6 ; R
2
3
(a)
R
RR
RR
1
1
2
2
ohm
63
63
8
9
1
so R 2 Ans
(b)
I
E
R
amp
12
2
so I 6 A Ans
(c)
I
1
1
E
R
amp
12
6
I
1
2 A Ans
I
2
2
E
R
12
3
I
2
4 A Ans
Worked Example 2.5
Q A 10 resistor, a 20 resistor and a 30 resistor are connected (a) in series, and then
(b) in parallel with each other. Calculate the total resistance for each of the two
connections.
A
R
1
10 ; R
2
2 0 ; R
3
3 0
(a) R R
1
R
2
R
3
ohm
10 20 30
so, R 6 0 Ans
(b)
111 1
1
RR R R
23
siemen
1
1
11
1
02030
0 0 05 0 033.. .
so,
R
1
1083.
5.46 Ans
40
Fundamental Electrical and Electronic Principles
Alternatively,
11
1
11
R
02030
632
60
11
60
S
so,
R
60
11
5.46 Ans
2.3 Potential Divider
When resistors are connected in series the p.d. developed across each
resistor will be in direct proportion to its resistance value. This is a
useful fact to bear in mind, since it means it is possible to calculate the
p.d.s without fi rst having to determine the circuit current. Consider two
resistors connected across a 50 V supply as shown in Fig. 2.7 . In order
to demonstrate the potential divider effect we will in this case fi rstly
calculate circuit current and hence the two p.d.s by applying Ohm’s law:
RR R
R
I
E
R
I
VIR
12
11
75 25 100
50
100
05
05 75
ohm
amp
A
volt
.
.
VV
VIR
V
1
22
2
37 5
05 25
12 5
.
.
.
V
volt
V
Ans
Ans
I
E
50 V
R
2
75 Ω
25 Ω
R
1
V
1
V
2
Fig. 2.7