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1 | 2990-2993 | This will tell us the potential difference, V = V (P) – V (N) = e – I r
[Eq (3 38) between the positive terminal P and the negative terminal N; I
here is the current flowing from N to P through the cell] If, while labelling
the current I through the cell one goes from P to N, then of course
V = e + I r
(3 |
1 | 2991-2994 | (3 38) between the positive terminal P and the negative terminal N; I
here is the current flowing from N to P through the cell] If, while labelling
the current I through the cell one goes from P to N, then of course
V = e + I r
(3 60)
(a)Having clarified labelling, we now state the rules and the proof:
Junction rule: At any junction, the sum of the currents entering
the junction is equal to the sum of currents leaving the junction
(Fig |
1 | 2992-2995 | 38) between the positive terminal P and the negative terminal N; I
here is the current flowing from N to P through the cell] If, while labelling
the current I through the cell one goes from P to N, then of course
V = e + I r
(3 60)
(a)Having clarified labelling, we now state the rules and the proof:
Junction rule: At any junction, the sum of the currents entering
the junction is equal to the sum of currents leaving the junction
(Fig 3 |
1 | 2993-2996 | If, while labelling
the current I through the cell one goes from P to N, then of course
V = e + I r
(3 60)
(a)Having clarified labelling, we now state the rules and the proof:
Junction rule: At any junction, the sum of the currents entering
the junction is equal to the sum of currents leaving the junction
(Fig 3 15) |
1 | 2994-2997 | 60)
(a)Having clarified labelling, we now state the rules and the proof:
Junction rule: At any junction, the sum of the currents entering
the junction is equal to the sum of currents leaving the junction
(Fig 3 15) Gustav Robert Kirchhoff
(1824 – 1887) German
physicist, professor at
Heidelberg
and
at
Berlin |
1 | 2995-2998 | 3 15) Gustav Robert Kirchhoff
(1824 – 1887) German
physicist, professor at
Heidelberg
and
at
Berlin Mainly known for
his
development
of
spectroscopy, he also
made many important
contributions to mathe-
matical physics, among
them, his first and
second rules for circuits |
1 | 2996-2999 | 15) Gustav Robert Kirchhoff
(1824 – 1887) German
physicist, professor at
Heidelberg
and
at
Berlin Mainly known for
his
development
of
spectroscopy, he also
made many important
contributions to mathe-
matical physics, among
them, his first and
second rules for circuits GUSTAV ROBERT KIRCHHOFF (1824 –
Rationalised 2023-24
Physics
98
EXAMPLE 3 |
1 | 2997-3000 | Gustav Robert Kirchhoff
(1824 – 1887) German
physicist, professor at
Heidelberg
and
at
Berlin Mainly known for
his
development
of
spectroscopy, he also
made many important
contributions to mathe-
matical physics, among
them, his first and
second rules for circuits GUSTAV ROBERT KIRCHHOFF (1824 –
Rationalised 2023-24
Physics
98
EXAMPLE 3 5
This applies equally well if instead of a junction of
several lines, we consider a point in a line |
1 | 2998-3001 | Mainly known for
his
development
of
spectroscopy, he also
made many important
contributions to mathe-
matical physics, among
them, his first and
second rules for circuits GUSTAV ROBERT KIRCHHOFF (1824 –
Rationalised 2023-24
Physics
98
EXAMPLE 3 5
This applies equally well if instead of a junction of
several lines, we consider a point in a line The proof of this rule follows from the fact that
when currents are steady, there is no accumulation
of charges at any junction or at any point in a line |
1 | 2999-3002 | GUSTAV ROBERT KIRCHHOFF (1824 –
Rationalised 2023-24
Physics
98
EXAMPLE 3 5
This applies equally well if instead of a junction of
several lines, we consider a point in a line The proof of this rule follows from the fact that
when currents are steady, there is no accumulation
of charges at any junction or at any point in a line Thus, the total current flowing in, (which is the rate
at which charge flows into the junction), must equal
the total current flowing out |
1 | 3000-3003 | 5
This applies equally well if instead of a junction of
several lines, we consider a point in a line The proof of this rule follows from the fact that
when currents are steady, there is no accumulation
of charges at any junction or at any point in a line Thus, the total current flowing in, (which is the rate
at which charge flows into the junction), must equal
the total current flowing out (b)
Loop rule: The algebraic sum of changes in
potential around any closed loop involving
resistors and cells in the loop is zero
(Fig |
1 | 3001-3004 | The proof of this rule follows from the fact that
when currents are steady, there is no accumulation
of charges at any junction or at any point in a line Thus, the total current flowing in, (which is the rate
at which charge flows into the junction), must equal
the total current flowing out (b)
Loop rule: The algebraic sum of changes in
potential around any closed loop involving
resistors and cells in the loop is zero
(Fig 3 |
1 | 3002-3005 | Thus, the total current flowing in, (which is the rate
at which charge flows into the junction), must equal
the total current flowing out (b)
Loop rule: The algebraic sum of changes in
potential around any closed loop involving
resistors and cells in the loop is zero
(Fig 3 15) |
1 | 3003-3006 | (b)
Loop rule: The algebraic sum of changes in
potential around any closed loop involving
resistors and cells in the loop is zero
(Fig 3 15) This rule is also obvious, since electric potential is
dependent on the location of the point |
1 | 3004-3007 | 3 15) This rule is also obvious, since electric potential is
dependent on the location of the point Thus
starting with any point if we come back to the same
point, the total change must be zero |
1 | 3005-3008 | 15) This rule is also obvious, since electric potential is
dependent on the location of the point Thus
starting with any point if we come back to the same
point, the total change must be zero In a closed
loop, we do come back to the starting point and
hence the rule |
1 | 3006-3009 | This rule is also obvious, since electric potential is
dependent on the location of the point Thus
starting with any point if we come back to the same
point, the total change must be zero In a closed
loop, we do come back to the starting point and
hence the rule FIGURE 3 |
1 | 3007-3010 | Thus
starting with any point if we come back to the same
point, the total change must be zero In a closed
loop, we do come back to the starting point and
hence the rule FIGURE 3 15 At junction a the current
leaving is I1 + I2 and current entering is I3 |
1 | 3008-3011 | In a closed
loop, we do come back to the starting point and
hence the rule FIGURE 3 15 At junction a the current
leaving is I1 + I2 and current entering is I3 The junction rule says I3 = I1 + I2 |
1 | 3009-3012 | FIGURE 3 15 At junction a the current
leaving is I1 + I2 and current entering is I3 The junction rule says I3 = I1 + I2 At point
h current entering is I1 |
1 | 3010-3013 | 15 At junction a the current
leaving is I1 + I2 and current entering is I3 The junction rule says I3 = I1 + I2 At point
h current entering is I1 There is only one
current leaving h and by junction rule
that will also be I1 |
1 | 3011-3014 | The junction rule says I3 = I1 + I2 At point
h current entering is I1 There is only one
current leaving h and by junction rule
that will also be I1 For the loops ‘ahdcba’
and ‘ahdefga’, the loop rules give –30I1 –
41 I3 + 45 = 0 and –30I1 + 21 I2 – 80 = 0 |
1 | 3012-3015 | At point
h current entering is I1 There is only one
current leaving h and by junction rule
that will also be I1 For the loops ‘ahdcba’
and ‘ahdefga’, the loop rules give –30I1 –
41 I3 + 45 = 0 and –30I1 + 21 I2 – 80 = 0 Example 3 |
1 | 3013-3016 | There is only one
current leaving h and by junction rule
that will also be I1 For the loops ‘ahdcba’
and ‘ahdefga’, the loop rules give –30I1 –
41 I3 + 45 = 0 and –30I1 + 21 I2 – 80 = 0 Example 3 5 A battery of 10 V and negligible internal resistance is
connected across the diagonally opposite corners of a cubical network
consisting of 12 resistors each of resistance 1 W (Fig |
1 | 3014-3017 | For the loops ‘ahdcba’
and ‘ahdefga’, the loop rules give –30I1 –
41 I3 + 45 = 0 and –30I1 + 21 I2 – 80 = 0 Example 3 5 A battery of 10 V and negligible internal resistance is
connected across the diagonally opposite corners of a cubical network
consisting of 12 resistors each of resistance 1 W (Fig 3 |
1 | 3015-3018 | Example 3 5 A battery of 10 V and negligible internal resistance is
connected across the diagonally opposite corners of a cubical network
consisting of 12 resistors each of resistance 1 W (Fig 3 16) |
1 | 3016-3019 | 5 A battery of 10 V and negligible internal resistance is
connected across the diagonally opposite corners of a cubical network
consisting of 12 resistors each of resistance 1 W (Fig 3 16) Determine
the equivalent resistance of the network and the current along each
edge of the cube |
1 | 3017-3020 | 3 16) Determine
the equivalent resistance of the network and the current along each
edge of the cube Z
FIGURE 3 |
1 | 3018-3021 | 16) Determine
the equivalent resistance of the network and the current along each
edge of the cube Z
FIGURE 3 16
Rationalised 2023-24
Current
Electricity
99
EXAMPLE 3 |
1 | 3019-3022 | Determine
the equivalent resistance of the network and the current along each
edge of the cube Z
FIGURE 3 16
Rationalised 2023-24
Current
Electricity
99
EXAMPLE 3 5
Similation for application of Kirchhoff’s rules:
http://www |
1 | 3020-3023 | Z
FIGURE 3 16
Rationalised 2023-24
Current
Electricity
99
EXAMPLE 3 5
Similation for application of Kirchhoff’s rules:
http://www phys |
1 | 3021-3024 | 16
Rationalised 2023-24
Current
Electricity
99
EXAMPLE 3 5
Similation for application of Kirchhoff’s rules:
http://www phys hawaii |
1 | 3022-3025 | 5
Similation for application of Kirchhoff’s rules:
http://www phys hawaii edu/~teb/optics/java/kirch3/
EXAMPLE 3 |
1 | 3023-3026 | phys hawaii edu/~teb/optics/java/kirch3/
EXAMPLE 3 6
Solution The network is not reducible to a simple series and parallel
combinations of resistors |
1 | 3024-3027 | hawaii edu/~teb/optics/java/kirch3/
EXAMPLE 3 6
Solution The network is not reducible to a simple series and parallel
combinations of resistors There is, however, a clear symmetry in the
problem which we can exploit to obtain the equivalent resistance of
the network |
1 | 3025-3028 | edu/~teb/optics/java/kirch3/
EXAMPLE 3 6
Solution The network is not reducible to a simple series and parallel
combinations of resistors There is, however, a clear symmetry in the
problem which we can exploit to obtain the equivalent resistance of
the network The paths AA¢, AD and AB are obviously symmetrically placed in the
network |
1 | 3026-3029 | 6
Solution The network is not reducible to a simple series and parallel
combinations of resistors There is, however, a clear symmetry in the
problem which we can exploit to obtain the equivalent resistance of
the network The paths AA¢, AD and AB are obviously symmetrically placed in the
network Thus, the current in each must be the same, say, I |
1 | 3027-3030 | There is, however, a clear symmetry in the
problem which we can exploit to obtain the equivalent resistance of
the network The paths AA¢, AD and AB are obviously symmetrically placed in the
network Thus, the current in each must be the same, say, I Further,
at the corners A¢, B and D, the incoming current I must split equally
into the two outgoing branches |
1 | 3028-3031 | The paths AA¢, AD and AB are obviously symmetrically placed in the
network Thus, the current in each must be the same, say, I Further,
at the corners A¢, B and D, the incoming current I must split equally
into the two outgoing branches In this manner, the current in all
the 12 edges of the cube are easily written down in terms of I, using
Kirchhoff’s first rule and the symmetry in the problem |
1 | 3029-3032 | Thus, the current in each must be the same, say, I Further,
at the corners A¢, B and D, the incoming current I must split equally
into the two outgoing branches In this manner, the current in all
the 12 edges of the cube are easily written down in terms of I, using
Kirchhoff’s first rule and the symmetry in the problem Next take a closed loop, say, ABCC¢EA, and apply Kirchhoff’s second
rule:
–IR – (1/2)IR – IR + e = 0
where R is the resistance of each edge and e the emf of battery |
1 | 3030-3033 | Further,
at the corners A¢, B and D, the incoming current I must split equally
into the two outgoing branches In this manner, the current in all
the 12 edges of the cube are easily written down in terms of I, using
Kirchhoff’s first rule and the symmetry in the problem Next take a closed loop, say, ABCC¢EA, and apply Kirchhoff’s second
rule:
–IR – (1/2)IR – IR + e = 0
where R is the resistance of each edge and e the emf of battery Thus,
e = 5
2 I R
The equivalent resistance Req of the network is
5
3
6
Req
R
=εI
=
For R = 1 W, Req = (5/6) W and for e = 10 V, the total current (= 3I) in
the network is
3I = 10 V/(5/6) W = 12 A, i |
1 | 3031-3034 | In this manner, the current in all
the 12 edges of the cube are easily written down in terms of I, using
Kirchhoff’s first rule and the symmetry in the problem Next take a closed loop, say, ABCC¢EA, and apply Kirchhoff’s second
rule:
–IR – (1/2)IR – IR + e = 0
where R is the resistance of each edge and e the emf of battery Thus,
e = 5
2 I R
The equivalent resistance Req of the network is
5
3
6
Req
R
=εI
=
For R = 1 W, Req = (5/6) W and for e = 10 V, the total current (= 3I) in
the network is
3I = 10 V/(5/6) W = 12 A, i e |
1 | 3032-3035 | Next take a closed loop, say, ABCC¢EA, and apply Kirchhoff’s second
rule:
–IR – (1/2)IR – IR + e = 0
where R is the resistance of each edge and e the emf of battery Thus,
e = 5
2 I R
The equivalent resistance Req of the network is
5
3
6
Req
R
=εI
=
For R = 1 W, Req = (5/6) W and for e = 10 V, the total current (= 3I) in
the network is
3I = 10 V/(5/6) W = 12 A, i e , I = 4 A
The current flowing in each edge can now be read off from the
Fig |
1 | 3033-3036 | Thus,
e = 5
2 I R
The equivalent resistance Req of the network is
5
3
6
Req
R
=εI
=
For R = 1 W, Req = (5/6) W and for e = 10 V, the total current (= 3I) in
the network is
3I = 10 V/(5/6) W = 12 A, i e , I = 4 A
The current flowing in each edge can now be read off from the
Fig 3 |
1 | 3034-3037 | e , I = 4 A
The current flowing in each edge can now be read off from the
Fig 3 16 |
1 | 3035-3038 | , I = 4 A
The current flowing in each edge can now be read off from the
Fig 3 16 It should be noted that because of the symmetry of the network, the
great power of Kirchhoff’s rules has not been very apparent in Example 3 |
1 | 3036-3039 | 3 16 It should be noted that because of the symmetry of the network, the
great power of Kirchhoff’s rules has not been very apparent in Example 3 5 |
1 | 3037-3040 | 16 It should be noted that because of the symmetry of the network, the
great power of Kirchhoff’s rules has not been very apparent in Example 3 5 In a general network, there will be no such simplification due to symmetry,
and only by application of Kirchhoff’s rules to junctions and closed loops
(as many as necessary to solve the unknowns in the network) can we
handle the problem |
1 | 3038-3041 | It should be noted that because of the symmetry of the network, the
great power of Kirchhoff’s rules has not been very apparent in Example 3 5 In a general network, there will be no such simplification due to symmetry,
and only by application of Kirchhoff’s rules to junctions and closed loops
(as many as necessary to solve the unknowns in the network) can we
handle the problem This will be illustrated in Example 3 |
1 | 3039-3042 | 5 In a general network, there will be no such simplification due to symmetry,
and only by application of Kirchhoff’s rules to junctions and closed loops
(as many as necessary to solve the unknowns in the network) can we
handle the problem This will be illustrated in Example 3 6 |
1 | 3040-3043 | In a general network, there will be no such simplification due to symmetry,
and only by application of Kirchhoff’s rules to junctions and closed loops
(as many as necessary to solve the unknowns in the network) can we
handle the problem This will be illustrated in Example 3 6 Example 3 |
1 | 3041-3044 | This will be illustrated in Example 3 6 Example 3 6 Determine the current in each branch of the network
shown in Fig |
1 | 3042-3045 | 6 Example 3 6 Determine the current in each branch of the network
shown in Fig 3 |
1 | 3043-3046 | Example 3 6 Determine the current in each branch of the network
shown in Fig 3 17 |
1 | 3044-3047 | 6 Determine the current in each branch of the network
shown in Fig 3 17 FIGURE 3 |
1 | 3045-3048 | 3 17 FIGURE 3 17
Rationalised 2023-24
Physics
100
EXAMPLE 3 |
1 | 3046-3049 | 17 FIGURE 3 17
Rationalised 2023-24
Physics
100
EXAMPLE 3 6
Solution Each branch of the network is assigned an unknown current
to be determined by the application of Kirchhoff’s rules |
1 | 3047-3050 | FIGURE 3 17
Rationalised 2023-24
Physics
100
EXAMPLE 3 6
Solution Each branch of the network is assigned an unknown current
to be determined by the application of Kirchhoff’s rules To reduce
the number of unknowns at the outset, the first rule of Kirchhoff is
used at every junction to assign the unknown current in each branch |
1 | 3048-3051 | 17
Rationalised 2023-24
Physics
100
EXAMPLE 3 6
Solution Each branch of the network is assigned an unknown current
to be determined by the application of Kirchhoff’s rules To reduce
the number of unknowns at the outset, the first rule of Kirchhoff is
used at every junction to assign the unknown current in each branch We then have three unknowns I1, I2 and I3 which can be found by
applying the second rule of Kirchhoff to three different closed loops |
1 | 3049-3052 | 6
Solution Each branch of the network is assigned an unknown current
to be determined by the application of Kirchhoff’s rules To reduce
the number of unknowns at the outset, the first rule of Kirchhoff is
used at every junction to assign the unknown current in each branch We then have three unknowns I1, I2 and I3 which can be found by
applying the second rule of Kirchhoff to three different closed loops Kirchhoff’s second rule for the closed loop ADCA gives,
10 – 4(I1– I2) + 2(I2 + I3 – I1) – I1 = 0
[3 |
1 | 3050-3053 | To reduce
the number of unknowns at the outset, the first rule of Kirchhoff is
used at every junction to assign the unknown current in each branch We then have three unknowns I1, I2 and I3 which can be found by
applying the second rule of Kirchhoff to three different closed loops Kirchhoff’s second rule for the closed loop ADCA gives,
10 – 4(I1– I2) + 2(I2 + I3 – I1) – I1 = 0
[3 61(a)]
that is, 7I1– 6I2 – 2I3 = 10
For the closed loop ABCA, we get
10 – 4I2– 2 (I2 + I3) – I1 = 0
that is, I1 + 6I2 + 2I3 =10
[3 |
1 | 3051-3054 | We then have three unknowns I1, I2 and I3 which can be found by
applying the second rule of Kirchhoff to three different closed loops Kirchhoff’s second rule for the closed loop ADCA gives,
10 – 4(I1– I2) + 2(I2 + I3 – I1) – I1 = 0
[3 61(a)]
that is, 7I1– 6I2 – 2I3 = 10
For the closed loop ABCA, we get
10 – 4I2– 2 (I2 + I3) – I1 = 0
that is, I1 + 6I2 + 2I3 =10
[3 61(b)]
For the closed loop BCDEB, we get
5 – 2 (I2 + I3) – 2 (I2 + I3 – I1) = 0
that is, 2I1 – 4I2 – 4I3 = –5
[3 |
1 | 3052-3055 | Kirchhoff’s second rule for the closed loop ADCA gives,
10 – 4(I1– I2) + 2(I2 + I3 – I1) – I1 = 0
[3 61(a)]
that is, 7I1– 6I2 – 2I3 = 10
For the closed loop ABCA, we get
10 – 4I2– 2 (I2 + I3) – I1 = 0
that is, I1 + 6I2 + 2I3 =10
[3 61(b)]
For the closed loop BCDEB, we get
5 – 2 (I2 + I3) – 2 (I2 + I3 – I1) = 0
that is, 2I1 – 4I2 – 4I3 = –5
[3 61(c)]
Equations (3 |
1 | 3053-3056 | 61(a)]
that is, 7I1– 6I2 – 2I3 = 10
For the closed loop ABCA, we get
10 – 4I2– 2 (I2 + I3) – I1 = 0
that is, I1 + 6I2 + 2I3 =10
[3 61(b)]
For the closed loop BCDEB, we get
5 – 2 (I2 + I3) – 2 (I2 + I3 – I1) = 0
that is, 2I1 – 4I2 – 4I3 = –5
[3 61(c)]
Equations (3 61 a, b, c) are three simultaneous equations in three
unknowns |
1 | 3054-3057 | 61(b)]
For the closed loop BCDEB, we get
5 – 2 (I2 + I3) – 2 (I2 + I3 – I1) = 0
that is, 2I1 – 4I2 – 4I3 = –5
[3 61(c)]
Equations (3 61 a, b, c) are three simultaneous equations in three
unknowns These can be solved by the usual method to give
I1 = 2 |
1 | 3055-3058 | 61(c)]
Equations (3 61 a, b, c) are three simultaneous equations in three
unknowns These can be solved by the usual method to give
I1 = 2 5A, I2 = 5
8 A, I3 = 7
18
A
The currents in the various branches of the network are
AB : 5
8 A, CA :
1
2 2 A, DEB :
7
1 8 A
AD :
7
18
A, CD : 0 A, BC :
1
2 2 A
It is easily verified that Kirchhoff’s second rule applied to the
remaining closed loops does not provide any additional independent
equation, that is, the above values of currents satisfy the second
rule for every closed loop of the network |
1 | 3056-3059 | 61 a, b, c) are three simultaneous equations in three
unknowns These can be solved by the usual method to give
I1 = 2 5A, I2 = 5
8 A, I3 = 7
18
A
The currents in the various branches of the network are
AB : 5
8 A, CA :
1
2 2 A, DEB :
7
1 8 A
AD :
7
18
A, CD : 0 A, BC :
1
2 2 A
It is easily verified that Kirchhoff’s second rule applied to the
remaining closed loops does not provide any additional independent
equation, that is, the above values of currents satisfy the second
rule for every closed loop of the network For example, the total voltage
drop over the closed loop BADEB
5
85
4
15
8
4
V
V
V
+
×
−
×
equal to zero, as required by Kirchhoff’s second rule |
1 | 3057-3060 | These can be solved by the usual method to give
I1 = 2 5A, I2 = 5
8 A, I3 = 7
18
A
The currents in the various branches of the network are
AB : 5
8 A, CA :
1
2 2 A, DEB :
7
1 8 A
AD :
7
18
A, CD : 0 A, BC :
1
2 2 A
It is easily verified that Kirchhoff’s second rule applied to the
remaining closed loops does not provide any additional independent
equation, that is, the above values of currents satisfy the second
rule for every closed loop of the network For example, the total voltage
drop over the closed loop BADEB
5
85
4
15
8
4
V
V
V
+
×
−
×
equal to zero, as required by Kirchhoff’s second rule 3 |
1 | 3058-3061 | 5A, I2 = 5
8 A, I3 = 7
18
A
The currents in the various branches of the network are
AB : 5
8 A, CA :
1
2 2 A, DEB :
7
1 8 A
AD :
7
18
A, CD : 0 A, BC :
1
2 2 A
It is easily verified that Kirchhoff’s second rule applied to the
remaining closed loops does not provide any additional independent
equation, that is, the above values of currents satisfy the second
rule for every closed loop of the network For example, the total voltage
drop over the closed loop BADEB
5
85
4
15
8
4
V
V
V
+
×
−
×
equal to zero, as required by Kirchhoff’s second rule 3 13 WHEATSTONE BRIDGE
As an application of Kirchhoff’s rules consider the circuit shown in
Fig |
1 | 3059-3062 | For example, the total voltage
drop over the closed loop BADEB
5
85
4
15
8
4
V
V
V
+
×
−
×
equal to zero, as required by Kirchhoff’s second rule 3 13 WHEATSTONE BRIDGE
As an application of Kirchhoff’s rules consider the circuit shown in
Fig 3 |
1 | 3060-3063 | 3 13 WHEATSTONE BRIDGE
As an application of Kirchhoff’s rules consider the circuit shown in
Fig 3 18, which is called the Wheatstone bridge |
1 | 3061-3064 | 13 WHEATSTONE BRIDGE
As an application of Kirchhoff’s rules consider the circuit shown in
Fig 3 18, which is called the Wheatstone bridge The bridge has
four resistors R1, R2, R3 and R4 |
1 | 3062-3065 | 3 18, which is called the Wheatstone bridge The bridge has
four resistors R1, R2, R3 and R4 Across one pair of diagonally opposite
points (A and C in the figure) a source is connected |
1 | 3063-3066 | 18, which is called the Wheatstone bridge The bridge has
four resistors R1, R2, R3 and R4 Across one pair of diagonally opposite
points (A and C in the figure) a source is connected This (i |
1 | 3064-3067 | The bridge has
four resistors R1, R2, R3 and R4 Across one pair of diagonally opposite
points (A and C in the figure) a source is connected This (i e |
1 | 3065-3068 | Across one pair of diagonally opposite
points (A and C in the figure) a source is connected This (i e , AC) is
called the battery arm |
1 | 3066-3069 | This (i e , AC) is
called the battery arm Between the other two vertices, B and D, a
galvanometer G (which is a device to detect currents) is connected |
1 | 3067-3070 | e , AC) is
called the battery arm Between the other two vertices, B and D, a
galvanometer G (which is a device to detect currents) is connected This
line, shown as BD in the figure, is called the galvanometer arm |
1 | 3068-3071 | , AC) is
called the battery arm Between the other two vertices, B and D, a
galvanometer G (which is a device to detect currents) is connected This
line, shown as BD in the figure, is called the galvanometer arm For simplicity, we assume that the cell has no internal resistance |
1 | 3069-3072 | Between the other two vertices, B and D, a
galvanometer G (which is a device to detect currents) is connected This
line, shown as BD in the figure, is called the galvanometer arm For simplicity, we assume that the cell has no internal resistance In
general there will be currents flowing across all the resistors as well as a
current Ig through G |
1 | 3070-3073 | This
line, shown as BD in the figure, is called the galvanometer arm For simplicity, we assume that the cell has no internal resistance In
general there will be currents flowing across all the resistors as well as a
current Ig through G Of special interest, is the case of a balanced bridge
where the resistors are such that Ig = 0 |
1 | 3071-3074 | For simplicity, we assume that the cell has no internal resistance In
general there will be currents flowing across all the resistors as well as a
current Ig through G Of special interest, is the case of a balanced bridge
where the resistors are such that Ig = 0 We can easily get the balance
condition, such that there is no current through G |
1 | 3072-3075 | In
general there will be currents flowing across all the resistors as well as a
current Ig through G Of special interest, is the case of a balanced bridge
where the resistors are such that Ig = 0 We can easily get the balance
condition, such that there is no current through G In this case, the
Kirchhoff’s junction rule applied to junctions D and B (see the figure)
Rationalised 2023-24
Current
Electricity
101
FIGURE 3 |
1 | 3073-3076 | Of special interest, is the case of a balanced bridge
where the resistors are such that Ig = 0 We can easily get the balance
condition, such that there is no current through G In this case, the
Kirchhoff’s junction rule applied to junctions D and B (see the figure)
Rationalised 2023-24
Current
Electricity
101
FIGURE 3 18
EXAMPLE 3 |
1 | 3074-3077 | We can easily get the balance
condition, such that there is no current through G In this case, the
Kirchhoff’s junction rule applied to junctions D and B (see the figure)
Rationalised 2023-24
Current
Electricity
101
FIGURE 3 18
EXAMPLE 3 7
immediately gives us the relations I1 = I3 and I2 = I4 |
1 | 3075-3078 | In this case, the
Kirchhoff’s junction rule applied to junctions D and B (see the figure)
Rationalised 2023-24
Current
Electricity
101
FIGURE 3 18
EXAMPLE 3 7
immediately gives us the relations I1 = I3 and I2 = I4 Next, we apply
Kirchhoff’s loop rule to closed loops ADBA and CBDC |
1 | 3076-3079 | 18
EXAMPLE 3 7
immediately gives us the relations I1 = I3 and I2 = I4 Next, we apply
Kirchhoff’s loop rule to closed loops ADBA and CBDC The first
loop gives
–I1 R1 + 0 + I2 R2 = 0 (Ig = 0)
(3 |
1 | 3077-3080 | 7
immediately gives us the relations I1 = I3 and I2 = I4 Next, we apply
Kirchhoff’s loop rule to closed loops ADBA and CBDC The first
loop gives
–I1 R1 + 0 + I2 R2 = 0 (Ig = 0)
(3 62)
and the second loop gives, upon using I3 = I1, I4 = I2
I2 R4 + 0 – I1 R3 = 0
(3 |
1 | 3078-3081 | Next, we apply
Kirchhoff’s loop rule to closed loops ADBA and CBDC The first
loop gives
–I1 R1 + 0 + I2 R2 = 0 (Ig = 0)
(3 62)
and the second loop gives, upon using I3 = I1, I4 = I2
I2 R4 + 0 – I1 R3 = 0
(3 63)
From Eq |
1 | 3079-3082 | The first
loop gives
–I1 R1 + 0 + I2 R2 = 0 (Ig = 0)
(3 62)
and the second loop gives, upon using I3 = I1, I4 = I2
I2 R4 + 0 – I1 R3 = 0
(3 63)
From Eq (3 |
1 | 3080-3083 | 62)
and the second loop gives, upon using I3 = I1, I4 = I2
I2 R4 + 0 – I1 R3 = 0
(3 63)
From Eq (3 62), we obtain,
1
2
2
1
I
R
I
R
=
whereas from Eq |
1 | 3081-3084 | 63)
From Eq (3 62), we obtain,
1
2
2
1
I
R
I
R
=
whereas from Eq (3 |
1 | 3082-3085 | (3 62), we obtain,
1
2
2
1
I
R
I
R
=
whereas from Eq (3 63), we obtain,
1
4
2
3
I
R
I
R
=
Hence, we obtain the condition
2
4
1
3
R
R
R
=R
[3 |
1 | 3083-3086 | 62), we obtain,
1
2
2
1
I
R
I
R
=
whereas from Eq (3 63), we obtain,
1
4
2
3
I
R
I
R
=
Hence, we obtain the condition
2
4
1
3
R
R
R
=R
[3 64(a)]
This last equation relating the four resistors is called the balance
condition for the galvanometer to give zero or null deflection |
1 | 3084-3087 | (3 63), we obtain,
1
4
2
3
I
R
I
R
=
Hence, we obtain the condition
2
4
1
3
R
R
R
=R
[3 64(a)]
This last equation relating the four resistors is called the balance
condition for the galvanometer to give zero or null deflection The Wheatstone bridge and its balance condition provide a practical
method for determination of an unknown resistance |
1 | 3085-3088 | 63), we obtain,
1
4
2
3
I
R
I
R
=
Hence, we obtain the condition
2
4
1
3
R
R
R
=R
[3 64(a)]
This last equation relating the four resistors is called the balance
condition for the galvanometer to give zero or null deflection The Wheatstone bridge and its balance condition provide a practical
method for determination of an unknown resistance Let us suppose we
have an unknown resistance, which we insert in the fourth arm; R4 is
thus not known |
1 | 3086-3089 | 64(a)]
This last equation relating the four resistors is called the balance
condition for the galvanometer to give zero or null deflection The Wheatstone bridge and its balance condition provide a practical
method for determination of an unknown resistance Let us suppose we
have an unknown resistance, which we insert in the fourth arm; R4 is
thus not known Keeping known resistances R1 and R2 in the first and
second arm of the bridge, we go on varying R3 till the galvanometer shows
a null deflection |
1 | 3087-3090 | The Wheatstone bridge and its balance condition provide a practical
method for determination of an unknown resistance Let us suppose we
have an unknown resistance, which we insert in the fourth arm; R4 is
thus not known Keeping known resistances R1 and R2 in the first and
second arm of the bridge, we go on varying R3 till the galvanometer shows
a null deflection The bridge then is balanced, and from the balance
condition the value of the unknown resistance R4 is given by,
2
4
3
1
R
R
=R R
[3 |
1 | 3088-3091 | Let us suppose we
have an unknown resistance, which we insert in the fourth arm; R4 is
thus not known Keeping known resistances R1 and R2 in the first and
second arm of the bridge, we go on varying R3 till the galvanometer shows
a null deflection The bridge then is balanced, and from the balance
condition the value of the unknown resistance R4 is given by,
2
4
3
1
R
R
=R R
[3 64(b)]
A practical device using this principle is called the meter bridge |
1 | 3089-3092 | Keeping known resistances R1 and R2 in the first and
second arm of the bridge, we go on varying R3 till the galvanometer shows
a null deflection The bridge then is balanced, and from the balance
condition the value of the unknown resistance R4 is given by,
2
4
3
1
R
R
=R R
[3 64(b)]
A practical device using this principle is called the meter bridge Example 3 |
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