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1 | 3090-3093 | 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 7 The four arms of a Wheatstone bridge (Fig |
1 | 3091-3094 | 64(b)]
A practical device using this principle is called the meter bridge Example 3 7 The four arms of a Wheatstone bridge (Fig 3 |
1 | 3092-3095 | Example 3 7 The four arms of a Wheatstone bridge (Fig 3 19) have
the following resistances:
AB = 100W, BC = 10W, CD = 5W, and DA = 60W |
1 | 3093-3096 | 7 The four arms of a Wheatstone bridge (Fig 3 19) have
the following resistances:
AB = 100W, BC = 10W, CD = 5W, and DA = 60W FIGURE 3 |
1 | 3094-3097 | 3 19) have
the following resistances:
AB = 100W, BC = 10W, CD = 5W, and DA = 60W FIGURE 3 19
Rationalised 2023-24
Physics
102
EXAMPLE 3 |
1 | 3095-3098 | 19) have
the following resistances:
AB = 100W, BC = 10W, CD = 5W, and DA = 60W FIGURE 3 19
Rationalised 2023-24
Physics
102
EXAMPLE 3 7
A galvanometer of 15W resistance is connected across BD |
1 | 3096-3099 | FIGURE 3 19
Rationalised 2023-24
Physics
102
EXAMPLE 3 7
A galvanometer of 15W resistance is connected across BD Calculate
the current through the galvanometer when a potential difference of
10 V is maintained across AC |
1 | 3097-3100 | 19
Rationalised 2023-24
Physics
102
EXAMPLE 3 7
A galvanometer of 15W resistance is connected across BD Calculate
the current through the galvanometer when a potential difference of
10 V is maintained across AC Solution Considering the mesh BADB, we have
100I1 + 15Ig – 60I2 = 0
or 20I1 + 3Ig – 12I2= 0
[3 |
1 | 3098-3101 | 7
A galvanometer of 15W resistance is connected across BD Calculate
the current through the galvanometer when a potential difference of
10 V is maintained across AC Solution Considering the mesh BADB, we have
100I1 + 15Ig – 60I2 = 0
or 20I1 + 3Ig – 12I2= 0
[3 65(a)]
Considering the mesh BCDB, we have
10 (I1 – Ig) – 15Ig – 5 (I2 + Ig) = 0
10I1 – 30Ig –5I2 = 0
2I1 – 6Ig – I2 = 0
[3 |
1 | 3099-3102 | Calculate
the current through the galvanometer when a potential difference of
10 V is maintained across AC Solution Considering the mesh BADB, we have
100I1 + 15Ig – 60I2 = 0
or 20I1 + 3Ig – 12I2= 0
[3 65(a)]
Considering the mesh BCDB, we have
10 (I1 – Ig) – 15Ig – 5 (I2 + Ig) = 0
10I1 – 30Ig –5I2 = 0
2I1 – 6Ig – I2 = 0
[3 65(b)]
Considering the mesh ADCEA,
60I2 + 5 (I2 + Ig) = 10
65I2 + 5Ig = 10
13I2 + Ig = 2
[3 |
1 | 3100-3103 | Solution Considering the mesh BADB, we have
100I1 + 15Ig – 60I2 = 0
or 20I1 + 3Ig – 12I2= 0
[3 65(a)]
Considering the mesh BCDB, we have
10 (I1 – Ig) – 15Ig – 5 (I2 + Ig) = 0
10I1 – 30Ig –5I2 = 0
2I1 – 6Ig – I2 = 0
[3 65(b)]
Considering the mesh ADCEA,
60I2 + 5 (I2 + Ig) = 10
65I2 + 5Ig = 10
13I2 + Ig = 2
[3 65(c)]
Multiplying Eq |
1 | 3101-3104 | 65(a)]
Considering the mesh BCDB, we have
10 (I1 – Ig) – 15Ig – 5 (I2 + Ig) = 0
10I1 – 30Ig –5I2 = 0
2I1 – 6Ig – I2 = 0
[3 65(b)]
Considering the mesh ADCEA,
60I2 + 5 (I2 + Ig) = 10
65I2 + 5Ig = 10
13I2 + Ig = 2
[3 65(c)]
Multiplying Eq (3 |
1 | 3102-3105 | 65(b)]
Considering the mesh ADCEA,
60I2 + 5 (I2 + Ig) = 10
65I2 + 5Ig = 10
13I2 + Ig = 2
[3 65(c)]
Multiplying Eq (3 65b) by 10
20I1 – 60Ig – 10I2 = 0
[3 |
1 | 3103-3106 | 65(c)]
Multiplying Eq (3 65b) by 10
20I1 – 60Ig – 10I2 = 0
[3 65(d)]
From Eqs |
1 | 3104-3107 | (3 65b) by 10
20I1 – 60Ig – 10I2 = 0
[3 65(d)]
From Eqs (3 |
1 | 3105-3108 | 65b) by 10
20I1 – 60Ig – 10I2 = 0
[3 65(d)]
From Eqs (3 65d) and (3 |
1 | 3106-3109 | 65(d)]
From Eqs (3 65d) and (3 65a) we have
63Ig – 2I2 = 0
I2 = 31 |
1 | 3107-3110 | (3 65d) and (3 65a) we have
63Ig – 2I2 = 0
I2 = 31 5Ig
[3 |
1 | 3108-3111 | 65d) and (3 65a) we have
63Ig – 2I2 = 0
I2 = 31 5Ig
[3 65(e)]
Substituting the value of I2 into Eq |
1 | 3109-3112 | 65a) we have
63Ig – 2I2 = 0
I2 = 31 5Ig
[3 65(e)]
Substituting the value of I2 into Eq [3 |
1 | 3110-3113 | 5Ig
[3 65(e)]
Substituting the value of I2 into Eq [3 65(c)], we get
13 (31 |
1 | 3111-3114 | 65(e)]
Substituting the value of I2 into Eq [3 65(c)], we get
13 (31 5Ig ) + Ig = 2
410 |
1 | 3112-3115 | [3 65(c)], we get
13 (31 5Ig ) + Ig = 2
410 5 Ig = 2
Ig = 4 |
1 | 3113-3116 | 65(c)], we get
13 (31 5Ig ) + Ig = 2
410 5 Ig = 2
Ig = 4 87 mA |
1 | 3114-3117 | 5Ig ) + Ig = 2
410 5 Ig = 2
Ig = 4 87 mA SUMMARY
1 |
1 | 3115-3118 | 5 Ig = 2
Ig = 4 87 mA SUMMARY
1 Current through a given area of a conductor is the net charge passing
per unit time through the area |
1 | 3116-3119 | 87 mA SUMMARY
1 Current through a given area of a conductor is the net charge passing
per unit time through the area 2 |
1 | 3117-3120 | SUMMARY
1 Current through a given area of a conductor is the net charge passing
per unit time through the area 2 To maintain a steady current, we must have a closed circuit in which
an external agency moves electric charge from lower to higher potential
energy |
1 | 3118-3121 | Current through a given area of a conductor is the net charge passing
per unit time through the area 2 To maintain a steady current, we must have a closed circuit in which
an external agency moves electric charge from lower to higher potential
energy The work done per unit charge by the source in taking the
charge from lower to higher potential energy (i |
1 | 3119-3122 | 2 To maintain a steady current, we must have a closed circuit in which
an external agency moves electric charge from lower to higher potential
energy The work done per unit charge by the source in taking the
charge from lower to higher potential energy (i e |
1 | 3120-3123 | To maintain a steady current, we must have a closed circuit in which
an external agency moves electric charge from lower to higher potential
energy The work done per unit charge by the source in taking the
charge from lower to higher potential energy (i e , from one terminal
of the source to the other) is called the electromotive force, or emf, of
the source |
1 | 3121-3124 | The work done per unit charge by the source in taking the
charge from lower to higher potential energy (i e , from one terminal
of the source to the other) is called the electromotive force, or emf, of
the source Note that the emf is not a force; it is the voltage difference
between the two terminals of a source in open circuit |
1 | 3122-3125 | e , from one terminal
of the source to the other) is called the electromotive force, or emf, of
the source Note that the emf is not a force; it is the voltage difference
between the two terminals of a source in open circuit 3 |
1 | 3123-3126 | , from one terminal
of the source to the other) is called the electromotive force, or emf, of
the source Note that the emf is not a force; it is the voltage difference
between the two terminals of a source in open circuit 3 Ohm’s law: The electric current I flowing through a substance is
proportional to the voltage V across its ends, i |
1 | 3124-3127 | Note that the emf is not a force; it is the voltage difference
between the two terminals of a source in open circuit 3 Ohm’s law: The electric current I flowing through a substance is
proportional to the voltage V across its ends, i e |
1 | 3125-3128 | 3 Ohm’s law: The electric current I flowing through a substance is
proportional to the voltage V across its ends, i e , V µ I or V = RI,
where R is called the resistance of the substance |
1 | 3126-3129 | Ohm’s law: The electric current I flowing through a substance is
proportional to the voltage V across its ends, i e , V µ I or V = RI,
where R is called the resistance of the substance The unit of resistance
is ohm: 1W = 1 V A–1 |
1 | 3127-3130 | e , V µ I or V = RI,
where R is called the resistance of the substance The unit of resistance
is ohm: 1W = 1 V A–1 Rationalised 2023-24
Current
Electricity
103
4 |
1 | 3128-3131 | , V µ I or V = RI,
where R is called the resistance of the substance The unit of resistance
is ohm: 1W = 1 V A–1 Rationalised 2023-24
Current
Electricity
103
4 The resistance R of a conductor depends on its length l and
cross-sectional area A through the relation,
l
R
ρA
=
where r, called resistivity is a property of the material and depends on
temperature and pressure |
1 | 3129-3132 | The unit of resistance
is ohm: 1W = 1 V A–1 Rationalised 2023-24
Current
Electricity
103
4 The resistance R of a conductor depends on its length l and
cross-sectional area A through the relation,
l
R
ρA
=
where r, called resistivity is a property of the material and depends on
temperature and pressure 5 |
1 | 3130-3133 | Rationalised 2023-24
Current
Electricity
103
4 The resistance R of a conductor depends on its length l and
cross-sectional area A through the relation,
l
R
ρA
=
where r, called resistivity is a property of the material and depends on
temperature and pressure 5 Electrical resistivity of substances varies over a very wide range |
1 | 3131-3134 | The resistance R of a conductor depends on its length l and
cross-sectional area A through the relation,
l
R
ρA
=
where r, called resistivity is a property of the material and depends on
temperature and pressure 5 Electrical resistivity of substances varies over a very wide range Metals
have low resistivity, in the range of 10–8 W m to 10–6 W m |
1 | 3132-3135 | 5 Electrical resistivity of substances varies over a very wide range Metals
have low resistivity, in the range of 10–8 W m to 10–6 W m Insulators
like glass and rubber have 1022 to 1024 times greater resistivity |
1 | 3133-3136 | Electrical resistivity of substances varies over a very wide range Metals
have low resistivity, in the range of 10–8 W m to 10–6 W m Insulators
like glass and rubber have 1022 to 1024 times greater resistivity Semiconductors like Si and Ge lie roughly in the middle range of
resistivity on a logarithmic scale |
1 | 3134-3137 | Metals
have low resistivity, in the range of 10–8 W m to 10–6 W m Insulators
like glass and rubber have 1022 to 1024 times greater resistivity Semiconductors like Si and Ge lie roughly in the middle range of
resistivity on a logarithmic scale 6 |
1 | 3135-3138 | Insulators
like glass and rubber have 1022 to 1024 times greater resistivity Semiconductors like Si and Ge lie roughly in the middle range of
resistivity on a logarithmic scale 6 In most substances, the carriers of current are electrons; in some
cases, for example, ionic crystals and electrolytic liquids, positive and
negative ions carry the electric current |
1 | 3136-3139 | Semiconductors like Si and Ge lie roughly in the middle range of
resistivity on a logarithmic scale 6 In most substances, the carriers of current are electrons; in some
cases, for example, ionic crystals and electrolytic liquids, positive and
negative ions carry the electric current 7 |
1 | 3137-3140 | 6 In most substances, the carriers of current are electrons; in some
cases, for example, ionic crystals and electrolytic liquids, positive and
negative ions carry the electric current 7 Current density j gives the amount of charge flowing per second per
unit area normal to the flow,
j = nq vd
where n is the number density (number per unit volume) of charge
carriers each of charge q, and vd is the drift velocity of the charge
carriers |
1 | 3138-3141 | In most substances, the carriers of current are electrons; in some
cases, for example, ionic crystals and electrolytic liquids, positive and
negative ions carry the electric current 7 Current density j gives the amount of charge flowing per second per
unit area normal to the flow,
j = nq vd
where n is the number density (number per unit volume) of charge
carriers each of charge q, and vd is the drift velocity of the charge
carriers For electrons q = – e |
1 | 3139-3142 | 7 Current density j gives the amount of charge flowing per second per
unit area normal to the flow,
j = nq vd
where n is the number density (number per unit volume) of charge
carriers each of charge q, and vd is the drift velocity of the charge
carriers For electrons q = – e If j is normal to a cross-sectional area
A and is constant over the area, the magnitude of the current I through
the area is nevd A |
1 | 3140-3143 | Current density j gives the amount of charge flowing per second per
unit area normal to the flow,
j = nq vd
where n is the number density (number per unit volume) of charge
carriers each of charge q, and vd is the drift velocity of the charge
carriers For electrons q = – e If j is normal to a cross-sectional area
A and is constant over the area, the magnitude of the current I through
the area is nevd A 8 |
1 | 3141-3144 | For electrons q = – e If j is normal to a cross-sectional area
A and is constant over the area, the magnitude of the current I through
the area is nevd A 8 Using E = V/l, I = nevd A, and Ohm’s law, one obtains
2
d
eE
ne v
m
ρm
=
The proportionality between the force eE on the electrons in a metal
due to the external field E and the drift velocity vd (not acceleration)
can be understood, if we assume that the electrons suffer collisions
with ions in the metal, which deflect them randomly |
1 | 3142-3145 | If j is normal to a cross-sectional area
A and is constant over the area, the magnitude of the current I through
the area is nevd A 8 Using E = V/l, I = nevd A, and Ohm’s law, one obtains
2
d
eE
ne v
m
ρm
=
The proportionality between the force eE on the electrons in a metal
due to the external field E and the drift velocity vd (not acceleration)
can be understood, if we assume that the electrons suffer collisions
with ions in the metal, which deflect them randomly If such collisions
occur on an average at a time interval t,
vd = at = eEt/m
where a is the acceleration of the electron |
1 | 3143-3146 | 8 Using E = V/l, I = nevd A, and Ohm’s law, one obtains
2
d
eE
ne v
m
ρm
=
The proportionality between the force eE on the electrons in a metal
due to the external field E and the drift velocity vd (not acceleration)
can be understood, if we assume that the electrons suffer collisions
with ions in the metal, which deflect them randomly If such collisions
occur on an average at a time interval t,
vd = at = eEt/m
where a is the acceleration of the electron This gives
2
m
ne
ρ
τ
=
9 |
1 | 3144-3147 | Using E = V/l, I = nevd A, and Ohm’s law, one obtains
2
d
eE
ne v
m
ρm
=
The proportionality between the force eE on the electrons in a metal
due to the external field E and the drift velocity vd (not acceleration)
can be understood, if we assume that the electrons suffer collisions
with ions in the metal, which deflect them randomly If such collisions
occur on an average at a time interval t,
vd = at = eEt/m
where a is the acceleration of the electron This gives
2
m
ne
ρ
τ
=
9 In the temperature range in which resistivity increases linearly with
temperature, the temperature coefficient of resistivity a is defined as
the fractional increase in resistivity per unit increase in temperature |
1 | 3145-3148 | If such collisions
occur on an average at a time interval t,
vd = at = eEt/m
where a is the acceleration of the electron This gives
2
m
ne
ρ
τ
=
9 In the temperature range in which resistivity increases linearly with
temperature, the temperature coefficient of resistivity a is defined as
the fractional increase in resistivity per unit increase in temperature 10 |
1 | 3146-3149 | This gives
2
m
ne
ρ
τ
=
9 In the temperature range in which resistivity increases linearly with
temperature, the temperature coefficient of resistivity a is defined as
the fractional increase in resistivity per unit increase in temperature 10 Ohm’s law is obeyed by many substances, but it is not a fundamental
law of nature |
1 | 3147-3150 | In the temperature range in which resistivity increases linearly with
temperature, the temperature coefficient of resistivity a is defined as
the fractional increase in resistivity per unit increase in temperature 10 Ohm’s law is obeyed by many substances, but it is not a fundamental
law of nature It fails if
(a) V depends on I non-linearly |
1 | 3148-3151 | 10 Ohm’s law is obeyed by many substances, but it is not a fundamental
law of nature It fails if
(a) V depends on I non-linearly (b) the relation between V and I depends on the sign of V for the same
absolute value of V |
1 | 3149-3152 | Ohm’s law is obeyed by many substances, but it is not a fundamental
law of nature It fails if
(a) V depends on I non-linearly (b) the relation between V and I depends on the sign of V for the same
absolute value of V (c)
The relation between V and I is non-unique |
1 | 3150-3153 | It fails if
(a) V depends on I non-linearly (b) the relation between V and I depends on the sign of V for the same
absolute value of V (c)
The relation between V and I is non-unique An example of (a) is when r increases with I (even if temperature is
kept fixed) |
1 | 3151-3154 | (b) the relation between V and I depends on the sign of V for the same
absolute value of V (c)
The relation between V and I is non-unique An example of (a) is when r increases with I (even if temperature is
kept fixed) A rectifier combines features (a) and (b) |
1 | 3152-3155 | (c)
The relation between V and I is non-unique An example of (a) is when r increases with I (even if temperature is
kept fixed) A rectifier combines features (a) and (b) GaAs shows the
feature (c) |
1 | 3153-3156 | An example of (a) is when r increases with I (even if temperature is
kept fixed) A rectifier combines features (a) and (b) GaAs shows the
feature (c) 11 |
1 | 3154-3157 | A rectifier combines features (a) and (b) GaAs shows the
feature (c) 11 When a source of emf e is connected to an external resistance R, the
voltage Vext across R is given by
Vext = IR =
R
R
r
ε
+
where r is the internal resistance of the source |
1 | 3155-3158 | GaAs shows the
feature (c) 11 When a source of emf e is connected to an external resistance R, the
voltage Vext across R is given by
Vext = IR =
R
R
r
ε
+
where r is the internal resistance of the source Rationalised 2023-24
Physics
104
Physical Quantity
Symbol
Dimensions
Unit
Remark
Electric current
I
[A]
A
SI base unit
Charge
Q, q
[T A]
C
Voltage, Electric
V
[M L
2 T
–3 A
–1]
V
Work/charge
potential difference
Electromotive force
e
[M L
2 T
–3 A
–1]
V
Work/charge
Resistance
R
[M L
2 T
–3 A
–2]
W
R = V/I
Resistivity
r
[M L
3 T
–3 A
–2]
W m
R = rl/A
Electrical
s
[M
–1 L
–3 T
3 A
2]
S
s = 1/r
conductivity
Electric field
E
[M L T
–3 A
–1]
V m
–1
Electric force
charge
Drift speed
vd
[L T
–1]
m s
–1
vd
e E
m
=
τ
Relaxation time
t
[T]
s
Current density
j
[L
–2 A]
A m
–2
current/area
Mobility
m
[M L
3 T
–4 A
–1]
m
2 V
–1s
–1
vd/
E
12 |
1 | 3156-3159 | 11 When a source of emf e is connected to an external resistance R, the
voltage Vext across R is given by
Vext = IR =
R
R
r
ε
+
where r is the internal resistance of the source Rationalised 2023-24
Physics
104
Physical Quantity
Symbol
Dimensions
Unit
Remark
Electric current
I
[A]
A
SI base unit
Charge
Q, q
[T A]
C
Voltage, Electric
V
[M L
2 T
–3 A
–1]
V
Work/charge
potential difference
Electromotive force
e
[M L
2 T
–3 A
–1]
V
Work/charge
Resistance
R
[M L
2 T
–3 A
–2]
W
R = V/I
Resistivity
r
[M L
3 T
–3 A
–2]
W m
R = rl/A
Electrical
s
[M
–1 L
–3 T
3 A
2]
S
s = 1/r
conductivity
Electric field
E
[M L T
–3 A
–1]
V m
–1
Electric force
charge
Drift speed
vd
[L T
–1]
m s
–1
vd
e E
m
=
τ
Relaxation time
t
[T]
s
Current density
j
[L
–2 A]
A m
–2
current/area
Mobility
m
[M L
3 T
–4 A
–1]
m
2 V
–1s
–1
vd/
E
12 Kirchhoff’s Rules –
(a) Junction Rule: At any junction of circuit elements, the sum of
currents entering the junction must equal the sum of currents
leaving it |
1 | 3157-3160 | When a source of emf e is connected to an external resistance R, the
voltage Vext across R is given by
Vext = IR =
R
R
r
ε
+
where r is the internal resistance of the source Rationalised 2023-24
Physics
104
Physical Quantity
Symbol
Dimensions
Unit
Remark
Electric current
I
[A]
A
SI base unit
Charge
Q, q
[T A]
C
Voltage, Electric
V
[M L
2 T
–3 A
–1]
V
Work/charge
potential difference
Electromotive force
e
[M L
2 T
–3 A
–1]
V
Work/charge
Resistance
R
[M L
2 T
–3 A
–2]
W
R = V/I
Resistivity
r
[M L
3 T
–3 A
–2]
W m
R = rl/A
Electrical
s
[M
–1 L
–3 T
3 A
2]
S
s = 1/r
conductivity
Electric field
E
[M L T
–3 A
–1]
V m
–1
Electric force
charge
Drift speed
vd
[L T
–1]
m s
–1
vd
e E
m
=
τ
Relaxation time
t
[T]
s
Current density
j
[L
–2 A]
A m
–2
current/area
Mobility
m
[M L
3 T
–4 A
–1]
m
2 V
–1s
–1
vd/
E
12 Kirchhoff’s Rules –
(a) Junction Rule: At any junction of circuit elements, the sum of
currents entering the junction must equal the sum of currents
leaving it (b) Loop Rule: The algebraic sum of changes in potential around any
closed loop must be zero |
1 | 3158-3161 | Rationalised 2023-24
Physics
104
Physical Quantity
Symbol
Dimensions
Unit
Remark
Electric current
I
[A]
A
SI base unit
Charge
Q, q
[T A]
C
Voltage, Electric
V
[M L
2 T
–3 A
–1]
V
Work/charge
potential difference
Electromotive force
e
[M L
2 T
–3 A
–1]
V
Work/charge
Resistance
R
[M L
2 T
–3 A
–2]
W
R = V/I
Resistivity
r
[M L
3 T
–3 A
–2]
W m
R = rl/A
Electrical
s
[M
–1 L
–3 T
3 A
2]
S
s = 1/r
conductivity
Electric field
E
[M L T
–3 A
–1]
V m
–1
Electric force
charge
Drift speed
vd
[L T
–1]
m s
–1
vd
e E
m
=
τ
Relaxation time
t
[T]
s
Current density
j
[L
–2 A]
A m
–2
current/area
Mobility
m
[M L
3 T
–4 A
–1]
m
2 V
–1s
–1
vd/
E
12 Kirchhoff’s Rules –
(a) Junction Rule: At any junction of circuit elements, the sum of
currents entering the junction must equal the sum of currents
leaving it (b) Loop Rule: The algebraic sum of changes in potential around any
closed loop must be zero 13 |
1 | 3159-3162 | Kirchhoff’s Rules –
(a) Junction Rule: At any junction of circuit elements, the sum of
currents entering the junction must equal the sum of currents
leaving it (b) Loop Rule: The algebraic sum of changes in potential around any
closed loop must be zero 13 The Wheatstone bridge is an arrangement of four resistances – R1, R2,
R3, R4 as shown in the text |
1 | 3160-3163 | (b) Loop Rule: The algebraic sum of changes in potential around any
closed loop must be zero 13 The Wheatstone bridge is an arrangement of four resistances – R1, R2,
R3, R4 as shown in the text The null-point condition is given by
3
1
2
4
R
RR
R
=
using which the value of one resistance can be determined, knowing
the other three resistances |
1 | 3161-3164 | 13 The Wheatstone bridge is an arrangement of four resistances – R1, R2,
R3, R4 as shown in the text The null-point condition is given by
3
1
2
4
R
RR
R
=
using which the value of one resistance can be determined, knowing
the other three resistances POINTS TO PONDER
1 |
1 | 3162-3165 | The Wheatstone bridge is an arrangement of four resistances – R1, R2,
R3, R4 as shown in the text The null-point condition is given by
3
1
2
4
R
RR
R
=
using which the value of one resistance can be determined, knowing
the other three resistances POINTS TO PONDER
1 Current is a scalar although we represent current with an arrow |
1 | 3163-3166 | The null-point condition is given by
3
1
2
4
R
RR
R
=
using which the value of one resistance can be determined, knowing
the other three resistances POINTS TO PONDER
1 Current is a scalar although we represent current with an arrow Currents do not obey the law of vector addition |
1 | 3164-3167 | POINTS TO PONDER
1 Current is a scalar although we represent current with an arrow Currents do not obey the law of vector addition That current is a
scalar also follows from it’s definition |
1 | 3165-3168 | Current is a scalar although we represent current with an arrow Currents do not obey the law of vector addition That current is a
scalar also follows from it’s definition The current I through an area
of cross-section is given by the scalar product of two vectors:
I = j |
1 | 3166-3169 | Currents do not obey the law of vector addition That current is a
scalar also follows from it’s definition The current I through an area
of cross-section is given by the scalar product of two vectors:
I = j DS
where j and DS are vectors |
1 | 3167-3170 | That current is a
scalar also follows from it’s definition The current I through an area
of cross-section is given by the scalar product of two vectors:
I = j DS
where j and DS are vectors Rationalised 2023-24
Current
Electricity
105
2 |
1 | 3168-3171 | The current I through an area
of cross-section is given by the scalar product of two vectors:
I = j DS
where j and DS are vectors Rationalised 2023-24
Current
Electricity
105
2 Refer to V-I curves of a resistor and a diode as drawn in the text |
1 | 3169-3172 | DS
where j and DS are vectors Rationalised 2023-24
Current
Electricity
105
2 Refer to V-I curves of a resistor and a diode as drawn in the text A
resistor obeys Ohm’s law while a diode does not |
1 | 3170-3173 | Rationalised 2023-24
Current
Electricity
105
2 Refer to V-I curves of a resistor and a diode as drawn in the text A
resistor obeys Ohm’s law while a diode does not The assertion that
V = IR is a statement of Ohm’s law is not true |
1 | 3171-3174 | Refer to V-I curves of a resistor and a diode as drawn in the text A
resistor obeys Ohm’s law while a diode does not The assertion that
V = IR is a statement of Ohm’s law is not true This equation defines
resistance and it may be applied to all conducting devices whether
they obey Ohm’s law or not |
1 | 3172-3175 | A
resistor obeys Ohm’s law while a diode does not The assertion that
V = IR is a statement of Ohm’s law is not true This equation defines
resistance and it may be applied to all conducting devices whether
they obey Ohm’s law or not The Ohm’s law asserts that the plot of I
versus V is linear i |
1 | 3173-3176 | The assertion that
V = IR is a statement of Ohm’s law is not true This equation defines
resistance and it may be applied to all conducting devices whether
they obey Ohm’s law or not The Ohm’s law asserts that the plot of I
versus V is linear i e |
1 | 3174-3177 | This equation defines
resistance and it may be applied to all conducting devices whether
they obey Ohm’s law or not The Ohm’s law asserts that the plot of I
versus V is linear i e , R is independent of V |
1 | 3175-3178 | The Ohm’s law asserts that the plot of I
versus V is linear i e , R is independent of V Equation E = r j leads to another statement of Ohm’s law, i |
1 | 3176-3179 | e , R is independent of V Equation E = r j leads to another statement of Ohm’s law, i e |
1 | 3177-3180 | , R is independent of V Equation E = r j leads to another statement of Ohm’s law, i e , a
conducting material obeys Ohm’s law when the resistivity of the
material does not depend on the magnitude and direction of applied
electric field |
1 | 3178-3181 | Equation E = r j leads to another statement of Ohm’s law, i e , a
conducting material obeys Ohm’s law when the resistivity of the
material does not depend on the magnitude and direction of applied
electric field 3 |
1 | 3179-3182 | e , a
conducting material obeys Ohm’s law when the resistivity of the
material does not depend on the magnitude and direction of applied
electric field 3 Homogeneous conductors like silver or semiconductors like pure
germanium or germanium containing impurities obey Ohm’s law within
some range of electric field values |
1 | 3180-3183 | , a
conducting material obeys Ohm’s law when the resistivity of the
material does not depend on the magnitude and direction of applied
electric field 3 Homogeneous conductors like silver or semiconductors like pure
germanium or germanium containing impurities obey Ohm’s law within
some range of electric field values If the field becomes too strong,
there are departures from Ohm’s law in all cases |
1 | 3181-3184 | 3 Homogeneous conductors like silver or semiconductors like pure
germanium or germanium containing impurities obey Ohm’s law within
some range of electric field values If the field becomes too strong,
there are departures from Ohm’s law in all cases 4 |
1 | 3182-3185 | Homogeneous conductors like silver or semiconductors like pure
germanium or germanium containing impurities obey Ohm’s law within
some range of electric field values If the field becomes too strong,
there are departures from Ohm’s law in all cases 4 Motion of conduction electrons in electric field E is the sum of (i)
motion due to random collisions and (ii) that due to E |
1 | 3183-3186 | If the field becomes too strong,
there are departures from Ohm’s law in all cases 4 Motion of conduction electrons in electric field E is the sum of (i)
motion due to random collisions and (ii) that due to E The motion
due to random collisions averages to zero and does not contribute to
vd (Chapter 11, Textbook of Class XI) |
1 | 3184-3187 | 4 Motion of conduction electrons in electric field E is the sum of (i)
motion due to random collisions and (ii) that due to E The motion
due to random collisions averages to zero and does not contribute to
vd (Chapter 11, Textbook of Class XI) vd , thus is only due to applied
electric field on the electron |
1 | 3185-3188 | Motion of conduction electrons in electric field E is the sum of (i)
motion due to random collisions and (ii) that due to E The motion
due to random collisions averages to zero and does not contribute to
vd (Chapter 11, Textbook of Class XI) vd , thus is only due to applied
electric field on the electron 5 |
1 | 3186-3189 | The motion
due to random collisions averages to zero and does not contribute to
vd (Chapter 11, Textbook of Class XI) vd , thus is only due to applied
electric field on the electron 5 The relation j = r v should be applied to each type of charge carriers
separately |
1 | 3187-3190 | vd , thus is only due to applied
electric field on the electron 5 The relation j = r v should be applied to each type of charge carriers
separately In a conducting wire, the total current and charge density
arises from both positive and negative charges:
j = r+ v+ + r– v–
rrrrr = r+ + r–
Now in a neutral wire carrying electric current,
rrrrr+ = – r–
Further, v+ ~ 0 which gives
rrrrr = 0
j = r– v
Thus, the relation j = r v does not apply to the total current charge
density |
1 | 3188-3191 | 5 The relation j = r v should be applied to each type of charge carriers
separately In a conducting wire, the total current and charge density
arises from both positive and negative charges:
j = r+ v+ + r– v–
rrrrr = r+ + r–
Now in a neutral wire carrying electric current,
rrrrr+ = – r–
Further, v+ ~ 0 which gives
rrrrr = 0
j = r– v
Thus, the relation j = r v does not apply to the total current charge
density 6 |
1 | 3189-3192 | The relation j = r v should be applied to each type of charge carriers
separately In a conducting wire, the total current and charge density
arises from both positive and negative charges:
j = r+ v+ + r– v–
rrrrr = r+ + r–
Now in a neutral wire carrying electric current,
rrrrr+ = – r–
Further, v+ ~ 0 which gives
rrrrr = 0
j = r– v
Thus, the relation j = r v does not apply to the total current charge
density 6 Kirchhoff’s junction rule is based on conservation of charge and the
outgoing currents add up and are equal to incoming current at a
junction |
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