fundamental electrical and electronic principles - dc circuits
DESCRIPTION
principles of dc circuitsTRANSCRIPT
D.C. Circuits
Chapter 2
Le
arn
ing
Ou
tco
me
s
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 law
s, and the concepts of potential
and current dividers.
This m
eans 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 sim
ple circuits, by applying Ohm ’ s law
.
2
Carry out the above calculations for more complex networks using Kirchhoff’s Law
s.
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 slidew
ire potentiometer.
31
Res
isto
rsca
sca
ded
or
connec
ted in
seri
es
or
2.1
R
esi
sto
rs in
Se
rie
s
When resistors are connected ‘ end-to-end ’ so that the same current
fl ows through them
all they are said to be ca
scad
ed or co
nn
ecte
d i
n
seri
es . Such a circuit is shown in Fig. 2.1 . Note that, for the sake of
simplicity, an ideal source of em
f has been used (no internal resistance).
E
V1
R1
R2
R3
V2
V3
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 m
ust equal the value of the applied
emf. Thus
VIR
VIR
VIR
11
22
33
!!
!volt;
volt;and
volt
However, the circuit current I depends ultim
ately on the applied emf E
and the total resistance
R offered by the circuit. Hence
EIR
EV
VV
! !"
"
volt.
Also,
volt
12
3
and substituting for E, V1 , V2 and V
3 in this last equation
we have
volt
IRIR
IRIR
!"
"1
23
and dividing this last equation by the common factor I
RR
RR
!"
"1
23ohm
(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.
Wo
rke
d E
xa
mp
le 2
.1
Q
F
or
the
cir
cuit
sh
ow
n in
Fig
. 2.2
ca
lcu
late
(a
) th
e c
ircu
it r
esi
sta
nce
, (b
) th
e c
ircu
it c
urr
en
t, (
c) t
he
p.d
.
de
ve
lop
ed
acr
oss
ea
ch r
esi
sto
r, a
nd
(d
) th
e p
ow
er
dis
sip
ate
d b
y t
he
co
mp
lete
cir
cuit
.
A
E !
24
V; R
1 !
33
0 !
; R 2
! 1
50
0 !
; R 3
! 4
70
!
E
V1
R1
R2
R3
V2
V3
I
33
0Ω
1.5
kΩ
47
0Ω
24
V
Fig
. 2.2
D.C. Circuits
33
(a)
R !
R 1
" R
2 "
R 3
oh
m
! 3
30
" 1
50
0 "
47
0
R
! 2
30
0 !
or
2.3
k !
An
s
(b)
I! !
E Ra
mp
24
23
00
I
! 1
0.4
3 m
A A
ns
(c)
V1 !
IR
1 vo
lt
! 1
0.4
3 $
10
% 3 $
33
0
V
1 !
3.4
4 V
An
s
V
2 !
IR
2 v
olt
! 1
0.4
3 $
10
% 3 $
15
00
V
2 !
15
.65
vo
lts
An
s
V
3 !
IR
3 v
olt
! 1
0.4
3 $
10
% 3 $
47
0
V
3 !
4.9
0 V
An
s
Note: T
he
su
m o
f th
e a
bo
ve p
.d.s
is 2
3.9
9 V
inst
ea
d o
f 2
4 V
du
e t
o t
he
ro
un
din
g
err
ors
in t
he
ca
lcu
lati
on
. It
sho
uld
als
o b
e n
ote
d t
ha
t th
e v
alu
e q
uo
ted
fo
r th
e
curr
en
t w
as 1
0.4
3 m
A w
he
rea
s th
e c
alc
ula
tor
an
swe
r is
10
.43
47
mA
. Th
is la
tte
r
va
lue
wa
s th
en
sto
red
in t
he
ca
lcu
lato
r m
em
ory
an
d u
sed
in t
he
ca
lcu
lati
on
s fo
r
pa
rt (
c), t
hu
s re
du
cin
g t
he
ro
un
din
g e
rro
rs t
o a
n a
cce
pta
ble
min
imu
m.
(d)
P !
EI
wa
tt
! 2
4 $
10
.43
$ 1
0 %
3
P
! 0
.25
W o
r 2
50
mW
An
s
It s
ho
uld
be
no
ted
th
at
the
po
we
r is
dis
sip
ate
d b
y th
e t
hre
e r
esi
sto
rs in
th
e
circ
uit
. He
nce
, th
e c
ircu
it p
ow
er
cou
ld h
av
e b
ee
n d
ete
rmin
ed
by
calc
ula
tin
g t
he
po
we
r d
issi
pa
ted
by
ea
ch o
f th
ese
an
d a
dd
ing
th
ese
va
lue
s to
giv
e t
he
to
tal.
Th
is
is s
ho
wn
be
low
, an
d s
erv
es,
as
a c
he
ck f
or
the
last
an
swe
r.
PR
P P11
1
11
11
1
! !$
$
! !$
$
% %
I2
32
23
2
04
30
33
0
35
93
04
30
5
wa
tt
mW
(.
)
.
(.
)00
0
63
33
04
30
47
0
58
33
2
! !$
$
!
%
1 11
11.
(.
)
.
mW
mW
P
tota
l p
ow
er:
w
att
so
mw
PP
PP
P
!"
"
!
12
3
25
04
4.
(N
ote
th
e w
ors
en
ing
eO
ect
of
con
tin
uo
us
rou
nd
ing
err
or)
34
Fundamental Electrical and Electronic Principles
Wo
rke
d E
xa
mp
le 2
.2
Q
T
wo
re
sist
ors
are
co
nn
ect
ed
in s
eri
es
acr
oss
a b
att
ery
of
em
f 1
2 V
. If
on
e o
f th
e r
esi
sto
rs h
as
a v
alu
e
of
16
! a
nd
it d
issi
pa
tes
a p
ow
er
of
4 W
, th
en
ca
lcu
late
(a
) th
e c
ircu
it c
urr
en
t, a
nd
(b
) th
e v
alu
e o
f th
e
oth
er
resi
sto
r.
A
Sin
ce t
he
on
ly t
wo
pie
ces
of
da
ta t
ha
t a
re d
ire
ctly
re
late
d t
o e
ach
oth
er
con
cern
the
16
! r
esi
sto
r a
nd
th
e p
ow
er
tha
t it
dis
sip
ate
s, t
he
n t
his
info
rma
tio
n m
ust
form
th
e s
tart
ing
po
int
for
the
so
luti
on
of
the
pro
ble
m. U
sin
g t
he
se d
ata
we
ca
n
de
term
ine
eit
he
r th
e c
urr
en
t th
rou
gh
or
the
p.d
. acr
oss
th
e 1
6 !
re
sist
or
(an
d it
is n
ot
imp
ort
an
t w
hic
h o
f th
ese
is c
alc
ula
ted
R r
st).
To
illu
stra
te t
his
po
int
bo
th
me
tho
ds
will
be
de
mo
nst
rate
d. T
he
ap
pro
pri
ate
cir
cuit
dia
gra
m, w
hic
h f
orm
s a
n
inte
gra
l pa
rt o
f th
e s
olu
tio
n, i
s sh
ow
n in
Fig
. 2.3
.
E
VA
B
AB
C1
6Ω
VB
C
I
12
V
Fig
. 2.3
E !
12
V; R BC !
16
! ; P BC !
4 W
(a)
I2 R BC !
P BC w
att
I2
!P RB
C BC
!!
4 60
.25
1
s
o I
! 0
.5 A
An
s
(b)
tota
l re
sist
an
ce, R
E!
Io
hm
!!
12
05.
24Ω
R AB !
R %
R BC
! 2
4 %
16
so R AB !
8 !
An
s
Alt
ern
ati
vely
, th
e p
rob
lem
ca
n b
e s
olv
ed
th
us:
(a)
V R
PBC
BC
BC
2
!w
att
VBC2
! P BC $
R BC !
4 $
16
! 6
4
D.C. Circuits
35
so
V BC !
8 V
I!
V R
BC
BC
am
p
!8 61
so I
! 0
.5 A
An
s
(b)
V AB !
E %
V BC v
olt
! 1
2 %
8
V AB !
4 V
RV
AB
AB
!I
!4 05.
so
R AB !
8 !
An
s
2.2
R
esi
sto
rs in
Pa
rall
el
When resistors are joined ‘ side-by-side ’ so that their corresponding ends
are connected together they are said to be connected in p
ara
llel . Using
this form
of connection m
eans 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 m
ay be analysed as follows:
Resis
tors
in
Pa
rall
el
or
or
or
I 1
I
I 2 I 3
R1
R2
R3
E
I
Fig
. 2.4
36
Fundamental Electrical and Electronic Principles
Since all three resistors are connected directly across the battery
term
inals then they all have the same voltage developed across them
.
In other words the voltage is the common factor in this arrangem
ent
of resistors. Now, each resistor will allow a certain value of current to
fl ow through it, depending upon its resistance value. Thus
IE R
IE R
IE R
11
22
33
!!
!amp;
amp;and
amp
The total circuit current I is determined by the applied emf and the total
circuit resistance
R ,
soamp
IE 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
soI
II
I!
""
12
3
and substituting the above expression for the currents:
E R
E R
E R
E R!
""
12
3
and dividing the above equation by the common factor E :
11
11
12
3R
RR
R!
""
siemen
(2.2
)
Note
: The above equation does NOT give the total resistance of the
circuit, but does give the total circuit c
on
du
ctan
ce ( G) which is measured
in Siemens (S). Thus, conductance is the reciprocal of resistance, so
to obtain the circuit resistance you m
ust then take the reciprocal of
the answ
er obtained from an equation of the form
of equation (2.2).
Co
nd
ucta
nce is
a m
easu
re o
f th
e ‘ w
illin
gnes
s ’ o
f a
mate
rial
or
circ
uit t
o al
low
curr
ent
to
fl ow
thro
ugh it
That is
siemen;and
ohm
11
RG
GR
!!
(2.3
)
However, when only two resistors are in parallel the combined
resistance m
ay be obtained directly by using the following equation:
RR
R
RR
!$ "
12
12
ohm
(2.4
)
D.C. Circuits
37
I 1 I 2 I 3
R1
R2
R3
E
I
1.5
kΩ
33
0Ω
47
0 Ω
24
V
Fig
. 2.5
In t
his
con
text
, the
wor
d
iden
tica
l m
eans
havi
ng t
he
sam
e va
lue
of r
esis
tance
A
E !
24
V; R
1 !
33
0 !
; R 2
! 1
50
0 !
; R 3
! 4
70
!
(a)
11
11
1R
RR
R!
""
23
sie
me
n
!"
"1
1
1
1
33
05
00
47
0
! 0
.00
30
3 "
0.0
00
66
7 "
0.0
021
3
! 0
.00
58
25
S
so
R !
171
.68
! A
ns
(re
cip
roca
l of
0.0
05
82
5)
(b)
I1
1
!E R
am
p
!2
4
33
0
I 1
! 7
2.7
3 m
A A
ns
I 22
!E R
am
p
!2
4
50
01
I 2
! 1
6 m
A A
ns
Wo
rke
d E
xa
mp
le 2
.3
Q
C
on
sid
eri
ng
th
e c
ircu
it o
f F
ig. 2
.5 , c
alc
ula
ted
(a
) th
e t
ota
l re
sist
an
ce o
f th
e c
ircu
it, (
b)
the
th
ree
bra
nch
curr
en
t, a
nd
(c)
th
e c
urr
en
t d
raw
n f
rom
th
e b
att
ery
.
If there are ‘x
’id
enti
cal resistors in parallel the total resistance is
simply
R/x
oh
ms.
38
Fundamental Electrical and Electronic Principles
I 33
!E R
am
p
!2
4
47
0
I 3
! 51
.06
mA
An
s
(c)
I
! I
1 "
I 2
" I
3 a
mp
! 7
2.7
3 "
16
" 51
.06
mA
so
I !
13
9.8
mA
An
s
Alt
ern
ati
vely
, th
e c
ircu
it c
urr
en
t co
uld
ha
ve
be
en
de
term
ine
d b
y u
sin
g t
he
va
lue
s fo
r E
an
d R a
s fo
llo
ws
I!
E Ra
mp
!2
4
76
81
1.
I
! 1
39
.8 m
A A
ns
Compare this exam
ple with worked exam
ple 2.1 (the same values
for the resistors and the em
f 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
arrangem
ent provides m
ore paths for current fl ow.
Wo
rke
d E
xa
mp
le 2
.4
Q
T
wo
re
sist
ors
, on
e o
f 6
! a
nd
th
e o
the
r o
f 3
! r
esi
sta
nce
, are
co
nn
ect
ed
in p
ara
lle
l acr
oss
a s
ou
rce
of
em
f o
f 1
2 V
. De
term
ine
(a
) th
e e
N e
ctiv
e r
esi
sta
nce
of
the
co
mb
ina
tio
n, (
b)
the
cu
rre
nt
dra
wn
fro
m t
he
sou
rce
, an
d (
c) t
he
cu
rre
nt
thro
ug
h e
ach
re
sist
or.
A
Th
e c
orr
esp
on
din
g c
ircu
it d
iag
ram
, su
ita
bly
lab
ell
ed
is s
ho
wn
in F
ig. 2
.6.
I 1
I
I 2
12
V
R2
R1
E3
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
oh
m
!$ "
!6
3
63
8 91
so
R !
2 !
An
s
(b)
I
!E R
am
p
!
12 2
so
I !
6 A
An
s
(c)
I1
1
!E R
am
p
!
12 6
I 1
! 2
A A
ns
I 22
!E R
!1
2 3
I 2
! 4
A A
ns
Wo
rke
d E
xa
mp
le 2
.5
Q
A
10
! r
esi
sto
r, a
20
! r
esi
sto
r a
nd
a 3
0 !
re
sist
or
are
co
nn
ect
ed
(a
) in
se
rie
s, a
nd
th
en
(b)
in p
ara
lle
l wit
h e
ach
oth
er.
Ca
lcu
late
th
e t
ota
l re
sist
an
ce f
or
ea
ch o
f th
e t
wo
con
ne
ctio
ns.
A
R 1
! 1
0 !
; R 2
! 2
0 !
; R 3
! 3
0 !
(a)
R
! R
1 "
R 2
" R
3 o
hm
! 1
0 "
20
" 3
0
so
, R !
60
! A
ns
(b)
1
11
1
1R
RR
R!
""
23
sie
me
n
!"
"!
""
1 1
11
10
20
30
00
05
00
33
..
.
so
, R
!1 1
08
3.
! 5
.46
! A
ns
40
Fundamental Electrical and Electronic Principles
A
lte
rna
tive
ly,
11 1
11
R!
""
!"
"
02
03
0
63
2
60
!11
60
S
so
, R!
60
11
! 5
.46
! A
ns
2.3
P
ote
nti
al D
ivid
er
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 m
ind, 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 dem
onstrate 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 IE R
I
VIR
!"
!"
!
! !!
! !$
12
11
75
25
100
50
100
05
05
75ohm
amp
A
volt
!
.
.
VV VIR
V
1 22
2
375
05
25
125
! ! !$
!
.
. .
V volt
V
An
s
An
s
I
E5
0 V
R2
75
Ω
25
Ω
R1
V1
V2
Fig
. 2.7
D.C. Circuits
41
Applying the potential divider technique, the two p.d.s may be obtained by
using the fact that the p.d. across a resistor is given by the ratio of its
resistance value to the total resistance of the circuit, expressed as a proportion
of the applied voltage. Although this sounds complicated it is very simple to
put into practice. Expressed in the form
of an equation it means
VR
RR
E1
1
12
!"
$volt
(2.5
)
and
VR
RR
E2
2
12
!"
$volt
(2.6
)
and using the above equations the p.d.s can m
ore sim
ply be calculated
as follows:
V V
1 2
75
100
50
375
25
100
50
125
!$
!
!$
!
. .
V
and
V
An
s
An
s
This technique is not restricted to only two resistors in series, but may
be applied to any number. For exam
ple, if there were three resistors in
series, then the p.d. across each m
ay be found from
VR
RR
RE
VR
RR
RE
VR
RR
RE
11
12
3
22
12
3
33
12
3
!"
"$
!"
"$
!"
"$
and
volt
2.4
C
urr
en
t D
ivid
er
It has been shown that when resistors are connected in parallel the total
circuit current divides between the alternative paths available. So far we
have determined the branch currents by calculating the common p.d.
across a parallel branch and dividing this by the respective resistance
values. However, these currents can be found directly, without the need to
calculate the branch p.d., by using the current divider theory. Consider
two resistors connected in parallel across a source of em
f 48 V as shown in
Fig. 2.8 . U
sing the p.d. m
ethod we can calculate the two currents as follow
s:
IE R
IE R
II
11
22
12
48
12
48
24
42
!!
!!
!!
and
amp
Aand
A
42
Fundamental Electrical and Electronic Principles
It is now worth noting the values of the resistors and the corresponding
currents. It is clear that
R1 is half the value of R2 . So, from the calculation
we obtain the quite logical result that
I 1 is twice the value of I 2 . That is,
a ratio of 2:1 applies in each case. Thus, the sm
aller resistor carries the
greater proportion of the total current. By stating the ratio as 2:1 we can
say that the current is split into three equal‘ parts ’ . Two ‘ parts ’ are fl owing
through one resistor and the remaining‘ part ’ through the other resistor.
Thus 2 3
$I fl ows through R
1
and 1 3
$I fl ows through R
2
Since I
! 6 A then
I I1 2
2 36
4
1 36
2
!$
!
!$
!
A A
In general we can say that I
R
RR
I1
2
12
!"
$(2
.7)
and
IR
RR
I2
1
12
!"
$(2
.8)
Note
: This is NOT the same ratio as for the potential divider. If you
compare (2.5) with (2.7) you will fi nd that the numerator in (2.5) is R
1
whereas in (2.7) the numerator is
R2 . There is a sim
ilar ‘ cross-over ’
when (2.6) and (2.8) are compared.
Again, the current divider theory is not limited to only two resistors in
parallel. Any number can be accommodated. However, with three or
I
I 2
R2
E2
4Ω
I 1
R1
12
Ω4
8 V
Fig
. 2.8
D.C. Circuits
43
more parallel resistors the current division m
ethod can be cumbersome
to use, and it is m
uch easier for mistakes to be made. For this reason it
is recommended that where more than two resistors exist in parallel the
‘ p.d. method ’ is used. This will be illustrated in the next section, but
for completeness the application to three resistors is shown below.
Consider the arrangem
ent shown in Fig. 2.9 :
11
11
1 3
1 4
1 6
43
2
12
12
3R
RR
R!
""
!"
"!
""
and exam
ining the numerator, we have 4 "
3 "
2 !
9 ‘ parts ’ .
I 1 I 2 I 3
R1
R2
R3
I
18
A
3Ω
6Ω
4Ω
Fig
. 2.9
Thus, the current ratios will be 4/9, 3/9 and 2/9 respectively for the
three resistors.
So,
A;
A;
AI
II
12
3
4 918
83 9
18
62 9
18
4!
$!
!$
!!
$!
2.5
S
eri
es/
Pa
rall
el C
om
bin
ati
on
s
Most practical circuits consist of resistors which are interconnected in
both series and parallel form
s. The simplest m
ethod of solving such a
circuit is to reduce the parallel branches to their equivalent resistance
values and hence reduce the circuit to a sim
ple series arrangem
ent.
This is best illustrated by m
eans of a worked exam
ple.
Wo
rke
d E
xa
mp
le 2
.6
Q
F
or
the
cir
cuit
sh
ow
n in
Fig
. 2.1
0 , c
alc
ula
te (
a)
the
cu
rre
nt
dra
wn
fro
m t
he
su
pp
ly, (
b)
the
cu
rre
nt
thro
ug
h t
he
6 !
re
sist
or,
an
d (
c) t
he
po
we
r d
issi
pa
ted
by
th
e 5
.6 !
re
sist
or.
A
Th
e R
rst
ste
p in
th
e s
olu
tio
n is
to
ske
tch
an
d la
be
l th
e c
ircu
it d
iag
ram
, cle
arl
y
sho
win
g a
ll c
urr
en
ts U
ow
ing
an
d id
en
tify
ing
ea
ch p
art
of
the
cir
cuit
as
sho
wn
in
44
Fundamental Electrical and Electronic Principles
Fig
. 2.11
. Als
o n
ote
th
at
sin
ce t
he
re is
no
me
nti
on
of
inte
rna
l re
sist
an
ce it
ma
y b
e
ass
um
ed
th
at
the
so
urc
e o
f e
mf
is id
ea
l.
5.6
Ω
6Ω
A I
I 2I 1
B
4Ω
C
R1
R2
E64 V
Fig
. 2.1
1
5.6
Ω2
.4Ω
VA
BV
BC
64
VE
AB
C
I
Fig
. 2.1
2
(a)
To d
ete
rmin
e t
he
cu
rre
nt
I d
raw
n f
rom
th
e b
att
ery
we
ne
ed
to
kn
ow
th
e
tota
l re
sist
an
ce R AC o
f th
e c
ircu
it.
RBC
!$ "
64
64
(u
sin
g
pro
du
ct
sum
fo
r tw
o r
esi
sto
rs in
pa
ralle
l))
!2
4 01
so R BC !
2.4
!
T
he
ori
gin
al c
ircu
it m
ay
no
w b
e r
ed
raw
n a
s in
Fig
2.1
2 .
6Ω
4Ω
5.6
Ω
64
V
E
Fig
. 2.1
0
D.C. Circuits
45
R AC !
R AB "
R BC o
hm
(re
sist
ors
in s
eri
es)
! 5
.6 "
2.4
so R AC !
8 !
I
!!
E
RAC
am
p6
4 8
so
I !
8 A
An
s
(b)
To R
nd
th
e c
urr
en
t I 1
th
rou
gh
th
e 6
! r
esi
sto
r w
e m
ay
use
eit
he
r o
f tw
o
me
tho
ds.
Bo
th o
f th
ese
are
no
w d
em
on
stra
ted
.
p.d. m
ethod:
V BC !
IR BC v
olt
( Fi
g. 2
.12
)
! 8
$ 2
.4
so, V BC !
19
.2 V
I 11
!V RB
Ca
mp
( Fig
. 2.11
)
!1
92
6.
so, I
1 !
3.2
A A
ns
T
his
an
swe
r m
ay
be
ch
eck
ed
as
foll
ow
s:
I 1!
V RBC 2
am
p
!!
19
2
44
8.
.A
a
nd
sin
ce I
! I
1 "
I 2
! 3
.2 "
4.8
! 8
A
w
hic
h a
gre
es
wit
h t
he
va
lue
fo
un
d in
(a
).
current division method:
C
on
sid
eri
ng
Fig
. 2.11
, th
e c
urr
en
t I
split
s in
to t
he
co
mp
on
en
ts I
1 a
nd
I 2
acc
ord
ing
to
th
e r
ati
o o
f th
e r
esi
sto
r v
alu
es.
However, y
ou
mu
st b
ea
r in
min
d
tha
t th
e larger
re
sist
or
carr
ies
the
smaller
pro
po
rtio
n o
f th
e t
ota
l cu
rre
nt.
II
1
1
!"
$R
RR
2
2
am
p
!"
$4
64
8
so, I
1 !
3.2
A A
ns
(c)
P A
B !
I 2 R AB w
att
! 8
2 $
5.6
so
, P AB !
35
8.4
W A
ns
A
lte
rna
tive
ly, P AB !
V AB I
wa
tt
w
he
re V AB !
E %
V BC v
olt
! 6
4 %
19
.2 !
44
.8 V
P AB !
44
.8 $
8
so, P AB !
35
8.4
W A
ns
46
Fundamental Electrical and Electronic Principles
A
Th
e R
rst
ste
p in
th
e s
olu
tio
n is
to
lab
el t
he
dia
gra
m c
lea
rly
wit
h le
tte
rs a
t th
e
jun
ctio
ns
an
d id
en
tify
ing
p.d
.s a
nd
bra
nch
cu
rre
nts
. Th
is s
ho
wn
in F
ig. 2
.14
.
R1
R3
R4
R5
R6
R2
4Ω
6Ω
3Ω
6Ω
8Ω
5Ω
E
18
V
VA
BV
BC
VC
D
I 5I 3
I 4
I 1 I 2I 6
AB
CD
I
Fig
. 2.1
4
Wo
rke
d E
xa
mp
le 2
.7
Q
F
or
the
cir
cuit
of
Fig
. 2.1
3 c
alc
ula
te (
a)
the
cu
rre
nt
dra
wn
fro
m t
he
so
urc
e, (
b)
the
p.d
. acr
oss
ea
ch
resi
sto
r, (
c) t
he
cu
rre
nt
thro
ug
h e
ach
re
sist
or,
an
d (
d)
the
po
we
r d
issi
pa
ted
by
th
e 5
! r
esi
sto
r.
R1
R3
R4
R5
R6
R2
4Ω
6Ω
3Ω
6Ω
8Ω
5Ω
E
18
V
Fig
. 2.1
3
(a)
RRR
RR
AB
!"
!$ "
!1
1
2
2
46
46
24
oh
m.
!
R BC !
5 !
1
11
11
11
RR
RR
CD
!"
"!
""
45
63
68
!"
"!
84
3
24
5
24
1S
R CD !
2
4 51
! 1
.6 !
R !
R AB "
R BC "
R CD o
hm
R !
2.4
" 5
" 1
.6 !
9 !
I!
!E R
am
p1
8 9
I !
2 A
An
s
D.C. Circuits
47
(b)
Th
e c
ircu
it h
as
be
en
re
du
ced
to
its
seri
es
eq
uiv
ale
nt
as
sho
wn
in F
ig. 2
.15
.
Usi
ng
th
is e
qu
iva
len
t ci
rcu
it it
is n
ow
a s
imp
le m
att
er
to c
alc
ula
te t
he
p.d
.
acr
oss
ea
ch s
ect
ion
of
the
cir
cuit
.
V AB !
IR AB v
olt
! 2
$ 2
.4
V AB !
4.8
V A
ns
(t
his
p.d
. is
com
mo
n t
o b
oth
R 1
an
d R
2 )
V BC !
IR BC v
olt
! 2
$ 5
V BC !
10
V A
ns
V CD !
IR CD v
olt
! 2
$ 1
.6
V CD
= 3
.2 V
An
s
(t
his
p.d
. is
com
mo
n t
o R
4 , R
5 a
nd
R 6 )
I
E
18
V
VA
BV
BC
VC
D
2.4
Ω5
Ω1
.6Ω
AD
BC
Fig
. 2.1
5
(c)
I 1 !
V RA
B 1
!4
8 4. o
r I
1 !
R
RR
2
2
6 10
21
"$
!$
I
I 1
! 1
.2 A
An
s I 1
! 1
.2 A
An
s
I 2 !
V RAB 2
48 6
!.
or
I 2
!
R
RR
1
11
"$
!$
2
4 02
I
I 2
! 0
.8 A
An
s I 2
! 0
.8 A
An
s
I 3 !
I !
2 A
An
s
I 4 !
V RCD 4
32 3
!.
or
11
11
RR
RR
CD
!"
"
45
6
I 4
! 1
.06
7 A
An
s !
""
11
1
36
8
I 5 !
V RC
D 5
32 6
!.
!
84
3
24
5
24
""
!1
I 5
! 0
.53
3 A
An
s so
I 4
!
8 52
1$
I 6
! V RC
D 6
32 8
!.
I 4
! 1
.06
7 A
An
s
I 6
! 0
.4 A
An
s I 5
!
4 52
1$
I 5 !
0.5
33
A A
ns
I 6 !
3 52
1$
I 6 !
0.4
A A
ns
48
Fundamental Electrical and Electronic Principles
N
oti
ce t
ha
t th
e p
.d. m
eth
od
is a
n e
asi
er
an
d le
ss c
um
be
rso
me
on
e t
ha
n
curr
en
t d
ivis
ion
wh
en
mo
re t
ha
n t
wo
re
sist
ors
are
co
nn
ect
ed
in p
ara
lle
l.
(d)
P 3
! I 3
2 R
3 w
att
or
V BC I 3
wa
tt
o
r V RB
C2 3
wa
tt
a
nd
usi
ng
th
e R
rst
of
the
se a
lte
rna
tive
eq
ua
tio
n:
P 3
! 2
2 $
5
P 3
! 2
0 W
An
s
It
is le
ft t
o t
he
re
ad
er
to c
on
R rm
th
at
the
oth
er
two
po
we
r e
qu
ati
on
s a
bo
ve
yie
ld t
he
sa
me
an
swe
r.
2.6
K
irch
ho
N ’ s
Cu
rre
nt
La
w
We have already put this law
into practice, though without stating it
explicitly. The law states that the algebraic sum of the currents at any
junction of a circuit is zero. Another, and perhaps simpler, way of
stating this is to say that the sum of the currents arriving at a junction
is equal to the sum of the currents leaving that junction. Thus we have
applied the law with parallel circuits, where the assumption has been
made that the sum of the branch currents equals the current drawn from
the source. Expressing the law in the form
of an equation we have:
&I
!0
(2.9
)
where the symbol # m
eans ‘ the sum of ’ .
Figure 2.16 illustrates a junction within a circuit with a number of currents
arriving and leaving the junction. Applying Kirchhoff ’ s current law yields:
II
II
I1
23
45
0%
""
%!
where ‘"
’ signs have been used to denote currents arriving and ‘
% ’
signs for currents leaving the junction. This equation can be transposed
to comply with the alternative statem
ent for the law, thus:
II
II
I1
34
25
""
!"
I 2
I 3
I 1
I 4
I 5
Fig
. 2.1
6
D.C. Circuits
49
Wo
rke
d E
xa
mp
le 2
.8
Q
F
or
the
ne
two
rk s
ho
wn
in F
ig. 2
.17
ca
lcu
late
th
e v
alu
es
of
the
ma
rke
d c
urr
en
ts.
40
A
10
A
80
A
30
A
25
A
A BC
D
I 2
I 1
I 5 I 4
I 3
F
E
Fig
. 2.1
7
A
Ju
nct
ion
A: I
2 !
40
" 1
0 !
50
A A
ns
Ju
nc
tio
n C
:
so
A
II
I
I
1 1
1
"!
"! !
28
0
50
80
30
An
s
Ju
nct
ion
D: I
3 !
80
" 3
0 !
11
0 A
An
s
Ju
nc
tio
n E
:
so
A
II
I I
43
4 4
25
02
5
85
"! !
%
!
11
An
s
Ju
nc
tio
n F
:
so
A
II
I
I I
54
5
5 5
30
85
30
30
85
55
"!
"! !
%
!%
An
s
Note: T
he
min
us
sig
n in
th
e la
st a
nsw
er
tell
s u
s th
at
the
cu
rre
nt I 5
is a
ctu
all
y
U o
win
g a
wa
y fr
om
th
e ju
nct
ion
ra
the
r th
an
to
wa
rds
it a
s sh
ow
n.
2.7
K
irch
ho
N ’ s
Vo
lta
ge
La
w
This law
also has already been used —
in the explanation of p.d. and in
the series and series/parallel circuits. This law
states that in any closed
network the algebraic sum of the em
fs is equal to the algebraic sum of
the p.d.s taken in order about the network. Once again, the law sounds
very complicated, but it is really only common sense, and is simple
to apply. So far, it has been applied only to very sim
ple circuits, such
as resistors connected in series across a source of em
f. In this case
we have said that the sum of the p.d.s is equal to the applied emf (e.g.
V1 "
V2 !
E ). However, these simple circuits have had only one source
50
Fundamental Electrical and Electronic Principles
of em
f, and could be solved using sim
ple Ohm ’ s law
techniques.
When m
ore than one source of em
f is involved, or the network is more
complex, then a network analysis method m
ust be used. Kirchhoff ’ s is
one of these methods.
Expressing the law in m
athem
atical form
:
&&
EIR
!(2
.10)
A generalised circuit requiring the application of Kirchhoff ’ s law
s is
shown in Fig. 2.18 . Note the following:
1
The circuit has been labelled with letters so that it is easy to refer
to a particular loop and the direction around the loop that is being
considered. Thus, if the left-hand loop is considered, and you wish
to trace a path around it in a clockwise direction, this would be
referred to as ABEFA. If a counterclockwise path was required, it
would be referred to as FEBAF or AFEBA.
FE
DCA
B
R2
E1
I 2I 1
R1
R3
(I1
" I
2)
E2
Fig
. 2.1
8
2
Current directions have been assumed and m
arked on the diagram.
As was found in the previous worked exam
ple (2.8), it may well
turn out that one or more of these currents actually fl ows in the
opposite direction to that m
arked. This result would be indicated by
a negative value obtained from the calculation. However, to ensure
consistency, make the initial assumption that all sources of em
f are
discharging current into the circuit; i.e. current leaves the positive
term
inal of each battery and enters at its negative term
inal. The
current law is also applied at this stage, which is why the current
fl owing through R
3 is marked as ( I1 "
I2 ) and not as I3 . This is an
important point since the solution involves the use of simultaneous
equations, and the fewer the number of ‘ unknowns ’ the simpler the
solution. Thus marking the third-branch current in this way m
eans
D.C. Circuits
51
that there are only two ‘ unknowns ’ to fi nd, nam
ely I1 and I2 . The
value for the third branch current, I3 , is then sim
ply found by using
the values obtained for I 1 and I2 .
3
If a negative value is obtained for a current then the minus sign
MUST be retained in any subsequent calculations. However, when
you quote the answ
er for such a current, m
ake a note to the effect
that it is fl owing in the opposite direction to that m
arked, e.g. from
C to D.
4
When tracing the path around a loop, concentrate solely on that
loop and ignore the remainder of the circuit. Also note that if you
are following the marked direction of current then the resulting
p.d.(s) are assigned positive values. If the direction of ‘ travel ’ is
opposite to the current arrow then the p.d. is assigned a negative
value.
Let us now apply these techniques to the circuit of Fig. 2.18 .
Consider fi rst the left-hand loop in a clockwise direction. Tracing
around the loop it can be seen that there is only one source of em
f
within it (nam
ely
E1 ). Thus the sum of the em
fs is simply E
1 volt.
Also, within the loop there are only two resistors ( R1 and R
2 ) which
will result in two p.d.s, I 1
R1 and ( I 1
" I2)R
3 volt. The resulting loop
equation will therefore be:
ABEFA: E
IR
II
R1
11
12
3!
""
()
[1]
Now taking the right-hand loop in a counterclockwise direction it can
be seen that again there is only one source of em
f and two resistors.
This results in the following loop equation:
CBEDC: E
IR
II
R2
22
12
3!
""
()
[2]
Finally, let us consider the loop around the edges of the diagram in a
clockwise direction. This follows the ‘ norm
al ’ direction for E1 but is
opposite to that forE2 , so the sum of the em
fs is E1
% E
2 volt. The loop
equation is therefore
ABCDEFA: E
EI
RI
R1
21
12
2%
!%
[3]
Since there are only two unknowns then only two sim
ultaneous
equations are required, and three have been written. However it is a
useful practice to do this as the‘ extra ’ equation m
ay contain m
ore
convenient numerical values for the coeffi cients of the ‘ unknown ’
currents.
The complete technique for the applications of Kirchhoff ’ s law
s
becomes clearer by the consideration of a worked exam
ple containing
numerical values.
52
Fundamental Electrical and Electronic Principles
Wo
rke
d E
xa
mp
le 2
.9
Q
F
or
the
cir
cuit
of
Fig
. 2.1
9 d
ete
rmin
e t
he
va
lue
an
d d
ire
ctio
n o
f th
e c
urr
en
t in
ea
ch b
ran
ch, a
nd
th
e p
.d.
acr
oss
th
e 1
0 !
re
sist
or.
10
Ω
2Ω
3Ω
10
V4
V
I 1
I 2
(I1
" I
2)
AB
C
FE
D
Fig
. 2.2
0
A
Th
e c
ircu
it is
R r
st la
be
lle
d a
nd
cu
rre
nt
U o
ws
ide
nti
R e
d a
nd
ma
rke
d b
y a
pp
lyin
g
the
cu
rre
nt
law
. Th
is is
sh
ow
n in
Fig
. 2.2
0 .
R1
E2
E1
R2
R3
10
Ω
2Ω
3Ω
10
V4
V
Fig
. 2.1
9
AB
EFA
:
1
1
1 1
04
32
63
2
2 2
%!
%
!%
II
II
so
……………
[]
AB
CD
EFA
:
11 1
11
1
1 1
1
03
0
30
10
03
02
22
22
2
!"
"
!"
"
!"
II
I
II
I
II(
)
[]
so
……………
D.C. Circuits
53
BC
DE
B :
42
0
20
10
40
23
22
22
2
!"
"
!"
"
!"
II
I
II
I
II
1 1
11
1 1
1
()
[]
so
……………
Insp
ect
ion
of
eq
ua
tio
ns
[1]
an
d [
2]
sho
ws
tha
t if
eq
ua
tio
n [1
] is
mu
ltip
lied
by
5
the
n t
he
co
e_
cie
nt
of
I 2 w
ill b
e t
he
sa
me
in b
oth
eq
ua
tio
ns.
Th
us,
if t
he
tw
o a
re
no
w a
dd
ed
th
en
th
e t
erm
co
nta
inin
g I
2 w
ill b
e e
limin
ate
d, a
nd
he
nce
a v
alu
e
can
be
ob
tain
ed
fo
r I 1
.
30
50
5
03
02
40
28
2 2
!%
$
!"
!
11
1
11
1
1 1 1
II
II
I
……………
……………
[]
[]
so I
1 !
4
0
28
! 1
.42
9 A
An
s
Su
bst
itu
tin
g t
his
va
lue
fo
r I 1
into
eq
ua
tio
n [
3]
yie
lds;
44
29
2
24
42
9
02
9
20
85
7
2
2 2
!"
!%
!%
!%
11
11
1 1.
.
..
I
I Iso
A
(c
ha
rge
) A
nns
()
..
.
()
.
II
II
11
"!
%!
!"
'
!$
2
22
42
90
85
70
57
2
05
72
A
vo
lt
An
s
VR
CD
CD
110
57
2so
V
VCD
!.
An
s
Wo
rke
d E
xa
mp
le 2
.10
Q
F
or
the
cir
cuit
sh
ow
n in
Fig
2.2
1 , u
se K
irch
ho
N ’ s
La
ws
to c
alc
ula
te (
a)
the
cu
rre
nt
V o
win
g in
ea
ch
bra
nch
of
the
cir
cuit
, an
d (
b)
the
p.d
. acr
oss
th
e 5
! r
esi
sto
r.
5 Ω
1.5
Ω2
Ω
6 V
4.5
V
Fig
. 2.2
1
54
Fundamental Electrical and Electronic Principles
A
Firs
tly
the
cir
cuit
is s
ketc
he
d a
nd
lab
ell
ed
an
d c
urr
en
ts id
en
tiR
ed
usi
ng
Kir
chh
oO
’ s c
urr
en
t la
w. T
his
is s
ho
wn
in F
ig. 2
.22
.
1.5
Ω
5Ω
R3
2Ω
R2
R1
(I1
" I
2)
6V
4.5
VE
1E
2
I 1I 2
AB
C
FE
D
Fig
. 2.2
2
(a)
We
ca
n n
ow
co
nsi
de
r th
ree
loo
ps
in t
he
cir
cuit
an
d w
rite
do
wn
th
e
corr
esp
on
din
g e
qu
ati
on
s u
sin
g K
irch
ho
O ’ s
vo
lta
ge
law
:
A
BE
FA
:
ER
R1
11
1
11
11
11
!"
"
!"
"!
""
!
II
I
II
II
II
()
.(
).
23
22
65
55
55
6
vo
lt
so,
665
52
.[
]I
I1
1"
……………
C
BE
DC
:
ER
R2
22
23
22
22
45
25
25
5
45
!"
"
!"
"!
""
!
II
I
II
II
II
()
.(
)
.
1 11
vo
lt
so,
557
22
II
1"
……………
[]
A
BC
DE
FA
:
EE
RR
11
1
1 1
1
11
%!
%
%!
%
!%
22
2 2 2
64
55
2
55
2
II
II
II
vo
lt
so,
..
..
[……………
33]
N
ow
, an
y p
air
of
the
se t
hre
e e
qu
ati
on
s m
ay
be
use
d t
o s
olv
e t
he
pro
ble
m,
usi
ng
th
e t
ech
niq
ue
of
sim
ult
an
eo
us
eq
ua
tio
ns.
We
sh
all
use
eq
ua
tio
ns
[1]
an
d [
3]
to e
limin
ate
th
e u
nk
no
wn
cu
rre
nt
I 2 , a
nd
he
nce
ob
tain
a v
alu
e f
or
curr
en
t I 1
. To
do
th
is w
e c
an
mu
ltip
ly [1
] b
y 2
an
d [
3]
by
5, a
nd
th
en
ad
d t
he
two
mo
diR
ed
eq
ua
tio
ns
tog
eth
er,
th
us:
11
11
1
1
1
1
23
02
75
75
03
5
95
20
52
2
!"
$
!%
$
!
II
II I……………
……………
..[
]
..
[]
..
11
1
11
he
nc
e,
A
I!
!9
5
20
50
95
. ..
An
s
D.C. Circuits
55
S
ub
stit
uti
ng
th
is v
alu
e f
or
I 1 in
to e
qu
ati
on
[3
] g
ive
s:
11
1
11 1
1
.(
..
)
.. .
..
55
09
52
54
27
2
24
27
50
07
2
2
2
!$
%
!%
!%
!%
I
I
Ih
en
ce
, 33
2
00
36
62
an
d
A
I!
%.
An
s
Note: T
he
min
us
sig
n in
th
e a
nsw
er
for
I 2 in
dic
ate
s th
at
this
cu
rre
nt
is
act
ua
lly U
ow
ing
in t
he
op
po
site
dir
ect
ion
to
th
at
ma
rke
d in
Fig
. 2.2
2 . T
his
me
an
s th
at
ba
tte
ry E
1 is
bo
th s
up
ply
ing
cu
rre
nt
to t
he
5 !
re
sist
or
and
charging
ba
tte
ry E
2 .
Cu
rre
nt
thro
ug
h 5
re
sist
or
am
p
so c
u
!!
"!
"%
II
11
20
95
00
36
6.
(.
)
rrre
nt
thro
ug
h 5
re
sist
or
A
!!
%!
09
50
03
66
09
4.
..
11
An
s
(b)
To o
bta
in t
he
p.d
. acr
oss
th
e 5
! r
esi
sto
r w
e c
an
eit
he
r su
btr
act
th
e p
.d.
(vo
lta
ge
dro
p)
acr
oss
R 1
fro
m t
he
em
f E 1
or
add
th
e p
.d. a
cro
ss R
2 to
em
f
E 2 , b
eca
use
E 2
is b
ein
g charged. A
th
ird
alt
ern
ati
ve is
to
mu
ltip
ly R
3 b
y th
e
curr
en
t U
ow
ing
th
rou
gh
it. A
ll t
hre
e m
eth
od
s w
ill b
e s
ho
wn
he
re, a
nd
,
pro
vid
ed
th
at
the
sa
me
an
swe
r is
ob
tain
ed
ea
ch t
ime
, th
e c
orr
ect
ne
ss o
f
the
an
swe
rs o
bta
ine
d in
pa
rt (
a)
will
be
co
nR
rme
d.
VE
R
V
BE
BE
!%
!%
$
!%
!
11
11
1
1I v
olt
so,
V
60
95
5
64
26
5
45
74
(.
.)
.
.A
ns
O
R:
v
olt
so,
VE
R
V
BE
BE
!
!
"
!
22
24
50
03
66
2
45
00
73
2
45
73I
.(
.)
..
.VV
An
s
O
R:
VR
V
BE
BE
!
"
()
.
.II
11
23
09
45
45
7
vo
lt
so,
V A
ns
T
he
ve
ry s
ma
ll d
i? e
ren
ces
be
twe
en
th
ese
th
ree
an
swe
rs is
du
e s
imp
ly
to r
ou
nd
ing
err
ors
, an
d s
o t
he
an
swe
rs t
o p
art
(a
) a
re v
eri
G e
d a
s
corr
ect
.
2.8
T
he
Wh
ea
tsto
ne
Bri
dg
e N
etw
ork
This is a network of interconnected resistors or other components,
depending on the application. Although the circuit contains only one
source of em
f, it requires the application of a network theorem such
as the Kirchhoff ’ s m
ethod for its solution. A typical network, suitably
labelled and with current fl ows identifi ed is shown in Fig. 2.23 .
56
Fund
amen
tal E
lectric
al and
Electronic Principles
Notice that although there are fi ve resistors, the current law has been
applied so as to minim
ise the number of‘ unknown ’ currents to three.
Thus only three simultaneous equations will be required for the
solution, though there are seven possible loops to choose from. These
seven loops are:
ABCDA; ADCA; ABDCA; ADBCA; ABDA; BCDB; and
ABCDA
If you trace around these loops you will fi nd that the last three do not
include the source of em
f, so for each of these loops the sum of the
emfs will be ZERO! Up to a point it doesn ’ t m
atter which three loops
are chosen provided that at least one of them
includes the source.
If you chose to use only the last three‘ zero emf ’ loops you would
succeed only in proving that zero equals zero!
The present level of study does not require you to solve simultaneous
equations containing three unknowns. It is nevertheless good practice
in the use of Kirchhoff’s law
s, and the seven equations for the above
loops are listed below. In order for you to gain this practice it is
suggested that you attem
pt this exercise before reading further, and
compare your results with those shown below.
ABCA :
E1
I1R1 !
( I 1 $
I3 ) R
3
ADCA :
E1
I2R2 !
( I 2
! I3 ) R
4
ABDCA :
E1
I1R1 !
I3R5 !
( I 2
! I3 ) R
4
ADBCA :
E1
I2R2
$ I3R5 !
( I 1
$ I3 ) R
3
ABDA :
0
I1R1 !
I3R5
$ I2R2
BCDB :
0
( I 1
$ I3 ) R
3$ ( I2 !
I3 ) R
4$
I 3R5
ABCDA :
0
I1R1 !
( I 1
$ I3 ) R
3$
( I 2
! I3 ) R
4$
I2R2
(I1
$ I
3)
(I2
! I
3)
R2
R4
R1
R5
R3
CA
I 1 I 2
I
E1I 3
DB
Fig
. 2.2
3
D.C. Circ
uits
57
As a check that the current law has been correctly applied, consider
junctions B and C:
current arriving at B
total current leaving
so
!
!
I II
II1
2
1II
II
II
II
II
II2
13
23
13
23
12
current arriving at C
$
!!
$
!!
!
()
()
I
Hence, current leaving battery
current returning to battery.
Wo
rke
d E
xa
mp
le 2
.11
Q
F
or
the
bri
dg
e n
etw
ork
sh
ow
n in
Fig
. 2.2
4 c
alc
ula
te t
he
cu
rre
nt
thro
ug
h e
ach
re
sist
or,
an
d t
he
cu
rre
nt
dra
wn
fro
m t
he
su
pp
ly.
(I1
$ I
3)
(I2
! I
3)
5Ω
3Ω
1Ω
6Ω
4Ω
CA
I 1 I 2
I
10 VI 3
DB
Fig
. 2.2
4
A
Th
e c
ircu
it is
G r
st la
be
lle
d a
nd
th
e c
urr
en
ts id
en
tiG
ed
usi
ng
th
e c
urr
en
t la
w a
s
sho
wn
in F
ig. 2
.24
.
A
BD
A :
06
53
06
35
13
2
23
!
$
$
!
II
I
II
I
1 1……………
[]
B
DC
B :
05
4
54
4
04
0
32
33
32
33
23
!
!$
$
!
!$
!
$
!!
II
II
I
II
II
I
II
I
1
1
1
1
1
()
()
………………
[]
2
58
Fund
amen
tal E
lectric
al and
Electronic Principles
A
DC
A :
11
1
03 3
04
3
22
3
22
3
23
!
!
!
!
!
II
I
II
I
II(
)
……………
[]
M
ult
iply
ing
eq
ua
tio
n [1
] b
y 2
, eq
ua
tio
n [
2]
by
3 a
nd
th
en
ad
din
g t
he
m
02
60
2
02
33
02
3
03
23
23
2
$
!"
$
!!
"
$
!
11
1
1
1 1II
I
II
I
I
……………
……………
[]
[]
440
43I……………
[]
M
ult
iply
ing
eq
ua
tio
n [
3]
by
3, e
qu
ati
on
[4
] b
y 4
an
d t
he
n a
dd
ing
th
em
30
23
33
02
60
44
30
63
30
23
23 3
3
!
"
$
!"
1 11 1
1
II
II I
I
……………
……………
[]
[]
663
08
4
.1 A
An
s
S
ub
stit
uti
ng
fo
r I 3
in e
qu
ati
on
[3
]
11
1 1
04
08
4
49
86
98
6
42
45
4
2
2 2
!
I
I I
.
. ..
A A
ns
S
ub
stit
uti
ng
fo
r I 3
an
d I
2 in
eq
ua
tio
n [
2]
04
24
54
84
44
29
4
42
94
40
74
$
!!
I
I I
1
1 1
1
1..
. ..
A A
ns
II
I
I
!
!
1 1
2
07
42
45
4
35
29
..
. A
An
s
S
ince
all
of
the
an
swe
rs o
bta
ine
d a
re p
osi
tive
va
lue
s th
en
th
e c
urr
en
ts w
ill
V o
w in
th
e d
ire
ctio
ns
ma
rke
d o
n t
he
cir
cuit
dia
gra
m.
Wo
rke
d E
xa
mp
le 2
.12
Q
If
th
e c
ircu
it o
f F
ig. 2
.24
is n
ow
am
en
de
d b
y s
imp
ly c
ha
ng
ing
th
e v
alu
e o
f R DC
fro
m 1
! t
o 2
! , c
alc
ula
te
the
cu
rre
nt
C o
win
g t
hro
ug
h t
he
5 !
re
sist
or
in t
he
ce
ntr
al l
imb
.
A
Th
e a
me
nd
ed
cir
cuit
dia
gra
m is
sh
ow
n in
Fig
. 2.2
5 .
A
BD
A :
06
53
06
35
32
23
!
$
$
!
II
I
II
I
1 11
……………
[]
D.C. Circ
uits
59
(I1
$ I
3)
(I2
! I
3)
5Ω
3Ω
2Ω
6Ω
4Ω
CA
I 1 I 2
10 VI 3
DB
Fig
. 2.2
5
B
DC
B :
05
24
52
24
4
04
2
32
33
32
33
23
!
!$
$
!
!$
!
$
!!
II
II
I
II
II
I
II
I
()
()
1 1
111………………
[]
2
M
ult
iply
ing
eq
ua
tio
n [1
] b
y 2
, eq
ua
tio
n [
2]
by
3 a
nd
ad
din
g t
he
m
02
60
2
02
63
32
3
04
3
23
23
3
$
!"
$
!!
"
11
1
1
1 1II
I
II
I
I
……………
……………
[]
[]
soo
A
I 30
A
ns
At fi rst sight this would seem to be a very odd result. Here we have a
resistor in the middle of a circuit with current being drawn from the
source, yet no current fl ows through this particular resistor! Now, in any
circuit, current will fl ow between two points only if there is a difference
of potential between the two points. So we must conclude that the
potentials at junctions B and D m
ust be the same. Since junction A is
a common point for both the 6 ! and 3 ! resistors, then the p.d. across
the 6 ! m
ust be the same as that across the 3 ! resistor. Sim
ilarly, since
point C is common to the 4 ! and 2 ! resistors, then the p.d. across
each of these must also be equal. This m
ay be verifi ed as follows.
Since I3 is zero then the 5 ! resistor plays no part in the circuit. In this
case we can ignore its presence and re-draw the circuit as in Fig. 2.26 .
Thus the circuit is reduced to a sim
ple series/parallel arrangem
ent that
can be analysed using sim
ple Ohm ’ s law
techniques.
RR
R
R
IE
R
I
VABC
ABC
ABC
AB
!
!
13
1 1
64
10 10
10
1
ohm
so
amp
so
A
!
"
IR
11
16
6 volt
V
60
Fund
amen
tal E
lectric
al and
Electronic Principles
Sim
ilarly,
A
and
V
R
I
VAD
C
AD
!
"
32
5
10 5
2
23
6
2
!
Thus V
AB
VAD
6 V, so the potentials at B and D are equal. In
this last exam
ple, the values of Rl , R2, R3 and R
4 are such to produce
what is known as the b
ala
nce c
on
dit
ion for the bridge. Being able to
produce this condition is what m
akes the bridge circuit such a useful
one for many applications in m
easurement system
s. The value of
resistance in the central lim
b has no effect on the balance conditions.
This is because, at balance, zero current fl ows through it. In addition,
the value of the em
f also has no effect on the balance conditions, but
will of course affect the values for I 1 and I2 . Consider the general case
of a bridge circuit as shown in Fig. 2.27 , where the values of resistors
R1 to R
4 are adjusted so that I3 is zero.
Ba
lan
ce c
on
dit
ion
ref
ers
to t
hat
conditio
n w
hen
zer
o
curr
ent
fl ow
s th
rough t
he
centr
al a
rm o
f th
e bri
dge
circ
uit, d
ue
to a
par
ticu
lar
com
bin
ation
of
resi
stor
valu
es in
the
four
‘ oute
r ’
arm
s of
the
bri
dge
(I1
$ I
3)
(I!
I3)
R5
R2
R4
R1
R3
CA
I 1 I 2
E
I 3
DB
Fig
. 2.2
7
I 1 I 2
3Ω
2Ω
6Ω
4Ω
CA
I 1R
1
R2
R3
R4
I 2
10 V
DB
Fig
. 2.2
6
D.C. Circ
uits
61
V
IR
VI
RAB
AD
11
22
and
but under the balance condition V
AB
VAD
so I1R1
I2R2 …
……
……
[1]
Sim
ilarly, V
BC
VD
C
so ( I 1
$I 3 )
& R
3
( I 2
! I3 ) R
4
but, I3
0, so current through R
3
I1
and current through R
4
I2 , therefore
I 1R3
I2R4 …
……
……
[2]
Dividing equation [1] by equation [2]:
IR
IR
IR
IR
R R
R R
11
13
22
24
1 3
2 4
so
This last equation m
ay be verifi ed by considering the values used in the
previous exam
ple where R1
6 ! , R2
3 ! , R3
4 ! and R
4
2 ! .
i.e. 6 3
4 2
So for balance, the ratio of the two resistors on the left-hand side of the
bridge equals the ratio of the two on the right-hand side.
However, a better way to express the balance condition in terms of the
resistor values is as follows. If the product of two diagonally opposite
resistors equals the product of the other pair of diagonally opposite
resistors, then the bridge is balanced, and zero current fl ows through
the central lim
b
i.e. R
RR
R1
42
3
(2.1
1)
and transposing equation (2.11) to m
ake R4 the subject we have
RR R
R4
2 13
(2
.12)
Thus if resistors R
1 ,R2 and R
3 can be set to known values, and adjusted
until a sensitive current measuring device inserted in the central lim
b
indicates zero current, then we have the basis for a sensitive resistance
measuring device.
62
Fund
amen
tal E
lectric
al and
Electronic Principles
Wo
rke
d E
xa
mp
le 2
.13
Q
A
Wh
ea
tsto
ne
Bri
dg
e t
yp
e c
ircu
it is
sh
ow
n in
Fig
. 2.2
8 . D
ete
rmin
e (
a)
the
p.d
. be
twe
en
te
rmin
als
B a
nd
D, a
nd
(b
) th
e v
alu
e t
o w
hic
h R 4
mu
st b
e a
dju
ste
d in
ord
er
to r
ed
uce
th
e c
urr
en
t th
rou
gh
R 3
to z
ero
(ba
lan
ce t
he
bri
dg
e).
8Ω
10
Ω2
Ω
R1
R2
R5
5Ω
CA
10
V
DB
R4
20
Ω
E
R3
Fig
. 2.2
8
A
Th
e c
ircu
it is
ske
tch
ed
an
d c
urr
en
ts m
ark
ed
, ap
ply
ing
Kir
chh
o?
’ s c
urr
en
t la
w, a
s
sho
wn
in F
ig. 2
.29
.
Kir
chh
o?
’ s v
olt
ag
e la
w is
no
w a
pp
lied
to
an
y th
ree
loo
ps.
No
te t
ha
t a
s in
th
is
case
th
ere
are
th
ree
un
kn
ow
ns
(I 1 ,
I 2 ,
an
d I 3 )
the
n w
e m
ust
ha
ve a
t le
ast
th
ree
eq
ua
tio
ns
in o
rde
r to
so
lve
th
e p
rob
lem
.
A
BD
A:
02
08
0
02
00
83
2
23
!
$
$
!
II
I
II
I
1 1
1
11
so,
………………
....
....
.[]
B
DC
B:
08
25
82
25
5
05
2
32
33
32
33
2
!
!$
$
!
!$
!
$
!!
II
II
I
II
II
I
II
()
()
1
1
1so
, 11
52
3I………………
....
....
[]
A
DC
A:
11 1
11
00
2
02
2
02
2
22
3
22
3
23
!
!
!
!
!
II
I
II
I
II(
)
....
.so
, ………………………
...[
]3
U
sin
g e
qu
ati
on
s [1
] a
nd
[2
] to
elim
ina
te I
1 w
e h
av
e:
02
00
8
02
08
60
24
23
23
$
!
$
!!
"
II
I
II
I
1
1
11
……………
…………
[]
[]
an
d a
dd
ing
,, 0
26
84
23
$
!I
I………………
...[
]
D.C. Circ
uits
63
a
nd
no
w u
sin
g e
qu
ati
on
s [3
] a
nd
[4
] w
e c
an
elim
ina
te I
2 a
s fo
llo
ws:
11 1
11
02
23
02
40
84
6
04
0
23
23
3
!
$
!"
II
II
I
……………
…………
[]
[]
an
d
A
vo
lt
so,
V
I
I
3
330
40
00
24
4
00
24
48
09
5
"
1 1 1
.
. .
VR
V
BD
BD
An
s
(b)
For
ba
lan
ce c
on
dit
ion
s
RR
RR
RRR
R
R
24
5
45
2
4
20
2
0
4
"
1 1 1
oh
m
so,
!
An
s
2.9
T
he
Wh
ea
tsto
ne
Bri
dg
e I
nst
rum
en
t
This is an instrument used for the accurate measurement of resistance
over a wide range of resistance values. It comprises three arms, the
resistances of which can be adjusted to known values. A fourth arm
contains the‘ unknown ’ resistance, and a central lim
b contains a sensitive
microam
meter (a galvanometer or ‘ galvo ’ ). The general arrangem
ent is
shown in Fig. 2.30 . Comparing this circuit with that of Fig. 2.27 and using
equation (2.12), the value of the resistance to be measured ( R
x ) is given by
RR R
Rx
m d
v
ohm
(I1
$ I
3)
(I2
! I
3)
10
Ω
2Ω
R1
R3
8Ω
R2
R5
5Ω
CA
I 1
I 2
10
V
I 3
DB
R4
20
Ω
E
Fig
. 2.2
9