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1 | 2790-2793 | 3)
=(75 3) 1 70 10
×
×
= 820 °C
that is, T2 = (820 + 27 0) °C = 847 °C
Thus, the steady temperature of the heating element (when heating
effect due to the current equals heat loss to the surroundings) is
847 °C |
1 | 2791-2794 | 3) 1 70 10
×
×
= 820 °C
that is, T2 = (820 + 27 0) °C = 847 °C
Thus, the steady temperature of the heating element (when heating
effect due to the current equals heat loss to the surroundings) is
847 °C Rationalised 2023-24
Physics
92
EXAMPLE 3 |
1 | 2792-2795 | 70 10
×
×
= 820 °C
that is, T2 = (820 + 27 0) °C = 847 °C
Thus, the steady temperature of the heating element (when heating
effect due to the current equals heat loss to the surroundings) is
847 °C Rationalised 2023-24
Physics
92
EXAMPLE 3 4
Example 3 |
1 | 2793-2796 | 0) °C = 847 °C
Thus, the steady temperature of the heating element (when heating
effect due to the current equals heat loss to the surroundings) is
847 °C Rationalised 2023-24
Physics
92
EXAMPLE 3 4
Example 3 4 The resistance of the platinum wire of a platinum
resistance thermometer at the ice point is 5 W and at steam point is
5 |
1 | 2794-2797 | Rationalised 2023-24
Physics
92
EXAMPLE 3 4
Example 3 4 The resistance of the platinum wire of a platinum
resistance thermometer at the ice point is 5 W and at steam point is
5 23 W |
1 | 2795-2798 | 4
Example 3 4 The resistance of the platinum wire of a platinum
resistance thermometer at the ice point is 5 W and at steam point is
5 23 W When the thermometer is inserted in a hot bath, the resistance
of the platinum wire is 5 |
1 | 2796-2799 | 4 The resistance of the platinum wire of a platinum
resistance thermometer at the ice point is 5 W and at steam point is
5 23 W When the thermometer is inserted in a hot bath, the resistance
of the platinum wire is 5 795 W |
1 | 2797-2800 | 23 W When the thermometer is inserted in a hot bath, the resistance
of the platinum wire is 5 795 W Calculate the temperature of the
bath |
1 | 2798-2801 | When the thermometer is inserted in a hot bath, the resistance
of the platinum wire is 5 795 W Calculate the temperature of the
bath Solution R0 = 5 W, R100 = 5 |
1 | 2799-2802 | 795 W Calculate the temperature of the
bath Solution R0 = 5 W, R100 = 5 23 W and Rt = 5 |
1 | 2800-2803 | Calculate the temperature of the
bath Solution R0 = 5 W, R100 = 5 23 W and Rt = 5 795 W
Now,
0
0
100
0
100,
(1
)
t
t
R
R
t
R
R
t
R
R
α
−
=
×
=
+
−
5 |
1 | 2801-2804 | Solution R0 = 5 W, R100 = 5 23 W and Rt = 5 795 W
Now,
0
0
100
0
100,
(1
)
t
t
R
R
t
R
R
t
R
R
α
−
=
×
=
+
−
5 795
5
100
5 |
1 | 2802-2805 | 23 W and Rt = 5 795 W
Now,
0
0
100
0
100,
(1
)
t
t
R
R
t
R
R
t
R
R
α
−
=
×
=
+
−
5 795
5
100
5 23
−5
=
×
−
= 0 |
1 | 2803-2806 | 795 W
Now,
0
0
100
0
100,
(1
)
t
t
R
R
t
R
R
t
R
R
α
−
=
×
=
+
−
5 795
5
100
5 23
−5
=
×
−
= 0 795
0 |
1 | 2804-2807 | 795
5
100
5 23
−5
=
×
−
= 0 795
0 23 ×100
= 345 |
1 | 2805-2808 | 23
−5
=
×
−
= 0 795
0 23 ×100
= 345 65 °C
3 |
1 | 2806-2809 | 795
0 23 ×100
= 345 65 °C
3 9 ELECTRICAL ENERGY, POWER
Consider a conductor with end points A and B, in which a current I is
flowing from A to B |
1 | 2807-2810 | 23 ×100
= 345 65 °C
3 9 ELECTRICAL ENERGY, POWER
Consider a conductor with end points A and B, in which a current I is
flowing from A to B The electric potential at A and B are denoted by V(A)
and V(B) respectively |
1 | 2808-2811 | 65 °C
3 9 ELECTRICAL ENERGY, POWER
Consider a conductor with end points A and B, in which a current I is
flowing from A to B The electric potential at A and B are denoted by V(A)
and V(B) respectively Since current is flowing from A to B, V(A) > V(B)
and the potential difference across AB is V = V(A) – V(B) > 0 |
1 | 2809-2812 | 9 ELECTRICAL ENERGY, POWER
Consider a conductor with end points A and B, in which a current I is
flowing from A to B The electric potential at A and B are denoted by V(A)
and V(B) respectively Since current is flowing from A to B, V(A) > V(B)
and the potential difference across AB is V = V(A) – V(B) > 0 In a time interval Dt, an amount of charge DQ = I Dt travels from A to
B |
1 | 2810-2813 | The electric potential at A and B are denoted by V(A)
and V(B) respectively Since current is flowing from A to B, V(A) > V(B)
and the potential difference across AB is V = V(A) – V(B) > 0 In a time interval Dt, an amount of charge DQ = I Dt travels from A to
B The potential energy of the charge at A, by definition, was Q V(A) and
similarly at B, it is Q V(B) |
1 | 2811-2814 | Since current is flowing from A to B, V(A) > V(B)
and the potential difference across AB is V = V(A) – V(B) > 0 In a time interval Dt, an amount of charge DQ = I Dt travels from A to
B The potential energy of the charge at A, by definition, was Q V(A) and
similarly at B, it is Q V(B) Thus, change in its potential energy DUpot is
DUpot = Final potential energy – Initial potential energy
= DQ[(V (B) – V (A)] = –DQ V
= –I VDt < 0
(3 |
1 | 2812-2815 | In a time interval Dt, an amount of charge DQ = I Dt travels from A to
B The potential energy of the charge at A, by definition, was Q V(A) and
similarly at B, it is Q V(B) Thus, change in its potential energy DUpot is
DUpot = Final potential energy – Initial potential energy
= DQ[(V (B) – V (A)] = –DQ V
= –I VDt < 0
(3 28)
If charges moved without collisions through the conductor, their
kinetic energy would also change so that the total energy is unchanged |
1 | 2813-2816 | The potential energy of the charge at A, by definition, was Q V(A) and
similarly at B, it is Q V(B) Thus, change in its potential energy DUpot is
DUpot = Final potential energy – Initial potential energy
= DQ[(V (B) – V (A)] = –DQ V
= –I VDt < 0
(3 28)
If charges moved without collisions through the conductor, their
kinetic energy would also change so that the total energy is unchanged Conservation of total energy would then imply that,
DK = –DUpot
(3 |
1 | 2814-2817 | Thus, change in its potential energy DUpot is
DUpot = Final potential energy – Initial potential energy
= DQ[(V (B) – V (A)] = –DQ V
= –I VDt < 0
(3 28)
If charges moved without collisions through the conductor, their
kinetic energy would also change so that the total energy is unchanged Conservation of total energy would then imply that,
DK = –DUpot
(3 29)
that is,
DK = I VDt > 0
(3 |
1 | 2815-2818 | 28)
If charges moved without collisions through the conductor, their
kinetic energy would also change so that the total energy is unchanged Conservation of total energy would then imply that,
DK = –DUpot
(3 29)
that is,
DK = I VDt > 0
(3 30)
Thus, in case charges were moving freely through the conductor under
the action of electric field, their kinetic energy would increase as they
move |
1 | 2816-2819 | Conservation of total energy would then imply that,
DK = –DUpot
(3 29)
that is,
DK = I VDt > 0
(3 30)
Thus, in case charges were moving freely through the conductor under
the action of electric field, their kinetic energy would increase as they
move We have, however, seen earlier that on the average, charge carriers
do not move with acceleration but with a steady drift velocity |
1 | 2817-2820 | 29)
that is,
DK = I VDt > 0
(3 30)
Thus, in case charges were moving freely through the conductor under
the action of electric field, their kinetic energy would increase as they
move We have, however, seen earlier that on the average, charge carriers
do not move with acceleration but with a steady drift velocity This is
because of the collisions with ions and atoms during transit |
1 | 2818-2821 | 30)
Thus, in case charges were moving freely through the conductor under
the action of electric field, their kinetic energy would increase as they
move We have, however, seen earlier that on the average, charge carriers
do not move with acceleration but with a steady drift velocity This is
because of the collisions with ions and atoms during transit During
collisions, the energy gained by the charges thus is shared with the atoms |
1 | 2819-2822 | We have, however, seen earlier that on the average, charge carriers
do not move with acceleration but with a steady drift velocity This is
because of the collisions with ions and atoms during transit During
collisions, the energy gained by the charges thus is shared with the atoms The atoms vibrate more vigorously, i |
1 | 2820-2823 | This is
because of the collisions with ions and atoms during transit During
collisions, the energy gained by the charges thus is shared with the atoms The atoms vibrate more vigorously, i e |
1 | 2821-2824 | During
collisions, the energy gained by the charges thus is shared with the atoms The atoms vibrate more vigorously, i e , the conductor heats up |
1 | 2822-2825 | The atoms vibrate more vigorously, i e , the conductor heats up Thus,
in an actual conductor, an amount of energy dissipated as heat in the
conductor during the time interval Dt is,
DW = I VDt
(3 |
1 | 2823-2826 | e , the conductor heats up Thus,
in an actual conductor, an amount of energy dissipated as heat in the
conductor during the time interval Dt is,
DW = I VDt
(3 31)
The energy dissipated per unit time is the power dissipated
P = DW/Dt and we have,
P = I V
(3 |
1 | 2824-2827 | , the conductor heats up Thus,
in an actual conductor, an amount of energy dissipated as heat in the
conductor during the time interval Dt is,
DW = I VDt
(3 31)
The energy dissipated per unit time is the power dissipated
P = DW/Dt and we have,
P = I V
(3 32)
Rationalised 2023-24
Current
Electricity
93
Using Ohm’s law V = IR, we get
P = I 2 R = V 2/R
(3 |
1 | 2825-2828 | Thus,
in an actual conductor, an amount of energy dissipated as heat in the
conductor during the time interval Dt is,
DW = I VDt
(3 31)
The energy dissipated per unit time is the power dissipated
P = DW/Dt and we have,
P = I V
(3 32)
Rationalised 2023-24
Current
Electricity
93
Using Ohm’s law V = IR, we get
P = I 2 R = V 2/R
(3 33)
as the power loss (“ohmic loss”) in a conductor of resistance R carrying a
current I |
1 | 2826-2829 | 31)
The energy dissipated per unit time is the power dissipated
P = DW/Dt and we have,
P = I V
(3 32)
Rationalised 2023-24
Current
Electricity
93
Using Ohm’s law V = IR, we get
P = I 2 R = V 2/R
(3 33)
as the power loss (“ohmic loss”) in a conductor of resistance R carrying a
current I It is this power which heats up, for example, the coil of an
electric bulb to incandescence, radiating out heat and light |
1 | 2827-2830 | 32)
Rationalised 2023-24
Current
Electricity
93
Using Ohm’s law V = IR, we get
P = I 2 R = V 2/R
(3 33)
as the power loss (“ohmic loss”) in a conductor of resistance R carrying a
current I It is this power which heats up, for example, the coil of an
electric bulb to incandescence, radiating out heat and light Where does the power come from |
1 | 2828-2831 | 33)
as the power loss (“ohmic loss”) in a conductor of resistance R carrying a
current I It is this power which heats up, for example, the coil of an
electric bulb to incandescence, radiating out heat and light Where does the power come from As we have
reasoned before, we need an external source to keep
a steady current through the conductor |
1 | 2829-2832 | It is this power which heats up, for example, the coil of an
electric bulb to incandescence, radiating out heat and light Where does the power come from As we have
reasoned before, we need an external source to keep
a steady current through the conductor It is clearly
this source which must supply this power |
1 | 2830-2833 | Where does the power come from As we have
reasoned before, we need an external source to keep
a steady current through the conductor It is clearly
this source which must supply this power In the
simple circuit shown with a cell (Fig |
1 | 2831-2834 | As we have
reasoned before, we need an external source to keep
a steady current through the conductor It is clearly
this source which must supply this power In the
simple circuit shown with a cell (Fig 3 |
1 | 2832-2835 | It is clearly
this source which must supply this power In the
simple circuit shown with a cell (Fig 3 11), it is the
chemical energy of the cell which supplies this power
for as long as it can |
1 | 2833-2836 | In the
simple circuit shown with a cell (Fig 3 11), it is the
chemical energy of the cell which supplies this power
for as long as it can The expressions for power, Eqs |
1 | 2834-2837 | 3 11), it is the
chemical energy of the cell which supplies this power
for as long as it can The expressions for power, Eqs (3 |
1 | 2835-2838 | 11), it is the
chemical energy of the cell which supplies this power
for as long as it can The expressions for power, Eqs (3 32) and (3 |
1 | 2836-2839 | The expressions for power, Eqs (3 32) and (3 33),
show the dependence of the power dissipated in a
resistor R on the current through it and the voltage
across it |
1 | 2837-2840 | (3 32) and (3 33),
show the dependence of the power dissipated in a
resistor R on the current through it and the voltage
across it Equation (3 |
1 | 2838-2841 | 32) and (3 33),
show the dependence of the power dissipated in a
resistor R on the current through it and the voltage
across it Equation (3 33) has an important application to
power transmission |
1 | 2839-2842 | 33),
show the dependence of the power dissipated in a
resistor R on the current through it and the voltage
across it Equation (3 33) has an important application to
power transmission Electrical power is transmitted
from power stations to homes and factories, which
may be hundreds of miles away, via transmission
cables |
1 | 2840-2843 | Equation (3 33) has an important application to
power transmission Electrical power is transmitted
from power stations to homes and factories, which
may be hundreds of miles away, via transmission
cables One obviously wants to minimise the power
loss in the transmission cables connecting the power stations to homes
and factories |
1 | 2841-2844 | 33) has an important application to
power transmission Electrical power is transmitted
from power stations to homes and factories, which
may be hundreds of miles away, via transmission
cables One obviously wants to minimise the power
loss in the transmission cables connecting the power stations to homes
and factories We shall see now how this can be achieved |
1 | 2842-2845 | Electrical power is transmitted
from power stations to homes and factories, which
may be hundreds of miles away, via transmission
cables One obviously wants to minimise the power
loss in the transmission cables connecting the power stations to homes
and factories We shall see now how this can be achieved Consider a
device R, to which a power P is to be delivered via transmission cables
having a resistance Rc to be dissipated by it finally |
1 | 2843-2846 | One obviously wants to minimise the power
loss in the transmission cables connecting the power stations to homes
and factories We shall see now how this can be achieved Consider a
device R, to which a power P is to be delivered via transmission cables
having a resistance Rc to be dissipated by it finally If V is the voltage
across R and I the current through it, then
P = V I
(3 |
1 | 2844-2847 | We shall see now how this can be achieved Consider a
device R, to which a power P is to be delivered via transmission cables
having a resistance Rc to be dissipated by it finally If V is the voltage
across R and I the current through it, then
P = V I
(3 34)
The connecting wires from the power station to the device has a finite
resistance Rc |
1 | 2845-2848 | Consider a
device R, to which a power P is to be delivered via transmission cables
having a resistance Rc to be dissipated by it finally If V is the voltage
across R and I the current through it, then
P = V I
(3 34)
The connecting wires from the power station to the device has a finite
resistance Rc The power dissipated in the connecting wires, which is
wasted is Pc with
Pc = I 2 Rc
2
2
c
P R
V
=
(3 |
1 | 2846-2849 | If V is the voltage
across R and I the current through it, then
P = V I
(3 34)
The connecting wires from the power station to the device has a finite
resistance Rc The power dissipated in the connecting wires, which is
wasted is Pc with
Pc = I 2 Rc
2
2
c
P R
V
=
(3 35)
from Eq |
1 | 2847-2850 | 34)
The connecting wires from the power station to the device has a finite
resistance Rc The power dissipated in the connecting wires, which is
wasted is Pc with
Pc = I 2 Rc
2
2
c
P R
V
=
(3 35)
from Eq (3 |
1 | 2848-2851 | The power dissipated in the connecting wires, which is
wasted is Pc with
Pc = I 2 Rc
2
2
c
P R
V
=
(3 35)
from Eq (3 32) |
1 | 2849-2852 | 35)
from Eq (3 32) Thus, to drive a device of power P, the power wasted in the
connecting wires is inversely proportional to V 2 |
1 | 2850-2853 | (3 32) Thus, to drive a device of power P, the power wasted in the
connecting wires is inversely proportional to V 2 The transmission cables
from power stations are hundreds of miles long and their resistance Rc is
considerable |
1 | 2851-2854 | 32) Thus, to drive a device of power P, the power wasted in the
connecting wires is inversely proportional to V 2 The transmission cables
from power stations are hundreds of miles long and their resistance Rc is
considerable To reduce Pc, these wires carry current at enormous values
of V and this is the reason for the high voltage danger signs on transmission
lines — a common sight as we move away from populated areas |
1 | 2852-2855 | Thus, to drive a device of power P, the power wasted in the
connecting wires is inversely proportional to V 2 The transmission cables
from power stations are hundreds of miles long and their resistance Rc is
considerable To reduce Pc, these wires carry current at enormous values
of V and this is the reason for the high voltage danger signs on transmission
lines — a common sight as we move away from populated areas Using
electricity at such voltages is not safe and hence at the other end, a device
called a transformer lowers the voltage to a value suitable for use |
1 | 2853-2856 | The transmission cables
from power stations are hundreds of miles long and their resistance Rc is
considerable To reduce Pc, these wires carry current at enormous values
of V and this is the reason for the high voltage danger signs on transmission
lines — a common sight as we move away from populated areas Using
electricity at such voltages is not safe and hence at the other end, a device
called a transformer lowers the voltage to a value suitable for use 3 |
1 | 2854-2857 | To reduce Pc, these wires carry current at enormous values
of V and this is the reason for the high voltage danger signs on transmission
lines — a common sight as we move away from populated areas Using
electricity at such voltages is not safe and hence at the other end, a device
called a transformer lowers the voltage to a value suitable for use 3 10 CELLS, EMF, INTERNAL RESISTANCE
We have already mentioned that a simple device to maintain a steady
current in an electric circuit is the electrolytic cell |
1 | 2855-2858 | Using
electricity at such voltages is not safe and hence at the other end, a device
called a transformer lowers the voltage to a value suitable for use 3 10 CELLS, EMF, INTERNAL RESISTANCE
We have already mentioned that a simple device to maintain a steady
current in an electric circuit is the electrolytic cell Basically a cell has
two electrodes, called the positive (P) and the negative (N), as shown in
FIGURE 3 |
1 | 2856-2859 | 3 10 CELLS, EMF, INTERNAL RESISTANCE
We have already mentioned that a simple device to maintain a steady
current in an electric circuit is the electrolytic cell Basically a cell has
two electrodes, called the positive (P) and the negative (N), as shown in
FIGURE 3 11 Heat is produced in the
resistor R which is connected across
the terminals of a cell |
1 | 2857-2860 | 10 CELLS, EMF, INTERNAL RESISTANCE
We have already mentioned that a simple device to maintain a steady
current in an electric circuit is the electrolytic cell Basically a cell has
two electrodes, called the positive (P) and the negative (N), as shown in
FIGURE 3 11 Heat is produced in the
resistor R which is connected across
the terminals of a cell The energy
dissipated in the resistor R comes from
the chemical energy of the electrolyte |
1 | 2858-2861 | Basically a cell has
two electrodes, called the positive (P) and the negative (N), as shown in
FIGURE 3 11 Heat is produced in the
resistor R which is connected across
the terminals of a cell The energy
dissipated in the resistor R comes from
the chemical energy of the electrolyte Rationalised 2023-24
Physics
94
Fig |
1 | 2859-2862 | 11 Heat is produced in the
resistor R which is connected across
the terminals of a cell The energy
dissipated in the resistor R comes from
the chemical energy of the electrolyte Rationalised 2023-24
Physics
94
Fig 3 |
1 | 2860-2863 | The energy
dissipated in the resistor R comes from
the chemical energy of the electrolyte Rationalised 2023-24
Physics
94
Fig 3 12 |
1 | 2861-2864 | Rationalised 2023-24
Physics
94
Fig 3 12 They are immersed in an electrolytic solution |
1 | 2862-2865 | 3 12 They are immersed in an electrolytic solution Dipped in
the solution, the electrodes exchange charges with the electrolyte |
1 | 2863-2866 | 12 They are immersed in an electrolytic solution Dipped in
the solution, the electrodes exchange charges with the electrolyte The positive electrode has a potential difference V+ (V+ > 0) between
itself and the electrolyte solution immediately adjacent to it marked
A in the figure |
1 | 2864-2867 | They are immersed in an electrolytic solution Dipped in
the solution, the electrodes exchange charges with the electrolyte The positive electrode has a potential difference V+ (V+ > 0) between
itself and the electrolyte solution immediately adjacent to it marked
A in the figure Similarly, the negative electrode develops a negative
potential – (V– ) (V– ≥ 0) relative to the electrolyte adjacent to it,
marked as B in the figure |
1 | 2865-2868 | Dipped in
the solution, the electrodes exchange charges with the electrolyte The positive electrode has a potential difference V+ (V+ > 0) between
itself and the electrolyte solution immediately adjacent to it marked
A in the figure Similarly, the negative electrode develops a negative
potential – (V– ) (V– ≥ 0) relative to the electrolyte adjacent to it,
marked as B in the figure When there is no current, the electrolyte
has the same potential throughout, so that the potential difference
between P and N is V+ – (–V–) = V+ + V– |
1 | 2866-2869 | The positive electrode has a potential difference V+ (V+ > 0) between
itself and the electrolyte solution immediately adjacent to it marked
A in the figure Similarly, the negative electrode develops a negative
potential – (V– ) (V– ≥ 0) relative to the electrolyte adjacent to it,
marked as B in the figure When there is no current, the electrolyte
has the same potential throughout, so that the potential difference
between P and N is V+ – (–V–) = V+ + V– This difference is called the
electromotive force (emf) of the cell and is denoted by e |
1 | 2867-2870 | Similarly, the negative electrode develops a negative
potential – (V– ) (V– ≥ 0) relative to the electrolyte adjacent to it,
marked as B in the figure When there is no current, the electrolyte
has the same potential throughout, so that the potential difference
between P and N is V+ – (–V–) = V+ + V– This difference is called the
electromotive force (emf) of the cell and is denoted by e Thus
e = V++V– > 0
(3 |
1 | 2868-2871 | When there is no current, the electrolyte
has the same potential throughout, so that the potential difference
between P and N is V+ – (–V–) = V+ + V– This difference is called the
electromotive force (emf) of the cell and is denoted by e Thus
e = V++V– > 0
(3 36)
Note that e is, actually, a potential difference and not a force |
1 | 2869-2872 | This difference is called the
electromotive force (emf) of the cell and is denoted by e Thus
e = V++V– > 0
(3 36)
Note that e is, actually, a potential difference and not a force The
name emf, however, is used because of historical reasons, and was
given at a time when the phenomenon was not understood properly |
1 | 2870-2873 | Thus
e = V++V– > 0
(3 36)
Note that e is, actually, a potential difference and not a force The
name emf, however, is used because of historical reasons, and was
given at a time when the phenomenon was not understood properly To understand the significance of e, consider a resistor R
connected across the cell (Fig |
1 | 2871-2874 | 36)
Note that e is, actually, a potential difference and not a force The
name emf, however, is used because of historical reasons, and was
given at a time when the phenomenon was not understood properly To understand the significance of e, consider a resistor R
connected across the cell (Fig 3 |
1 | 2872-2875 | The
name emf, however, is used because of historical reasons, and was
given at a time when the phenomenon was not understood properly To understand the significance of e, consider a resistor R
connected across the cell (Fig 3 12) |
1 | 2873-2876 | To understand the significance of e, consider a resistor R
connected across the cell (Fig 3 12) A current I flows across R
from C to D |
1 | 2874-2877 | 3 12) A current I flows across R
from C to D As explained before, a steady current is maintained
because current flows from N to P through the electrolyte |
1 | 2875-2878 | 12) A current I flows across R
from C to D As explained before, a steady current is maintained
because current flows from N to P through the electrolyte Clearly,
across the electrolyte the same current flows through the electrolyte
but from N to P, whereas through R, it flows from P to N |
1 | 2876-2879 | A current I flows across R
from C to D As explained before, a steady current is maintained
because current flows from N to P through the electrolyte Clearly,
across the electrolyte the same current flows through the electrolyte
but from N to P, whereas through R, it flows from P to N The electrolyte through which a current flows has a finite
resistance r, called the internal resistance |
1 | 2877-2880 | As explained before, a steady current is maintained
because current flows from N to P through the electrolyte Clearly,
across the electrolyte the same current flows through the electrolyte
but from N to P, whereas through R, it flows from P to N The electrolyte through which a current flows has a finite
resistance r, called the internal resistance Consider first the
situation when R is infinite so that I = V/R = 0, where V is the
potential difference between P and N |
1 | 2878-2881 | Clearly,
across the electrolyte the same current flows through the electrolyte
but from N to P, whereas through R, it flows from P to N The electrolyte through which a current flows has a finite
resistance r, called the internal resistance Consider first the
situation when R is infinite so that I = V/R = 0, where V is the
potential difference between P and N Now,
V = Potential difference between P and A
+ Potential difference between A and B
+ Potential difference between B and N
= e
(3 |
1 | 2879-2882 | The electrolyte through which a current flows has a finite
resistance r, called the internal resistance Consider first the
situation when R is infinite so that I = V/R = 0, where V is the
potential difference between P and N Now,
V = Potential difference between P and A
+ Potential difference between A and B
+ Potential difference between B and N
= e
(3 37)
Thus, emf e is the potential difference between the positive and
negative electrodes in an open circuit, i |
1 | 2880-2883 | Consider first the
situation when R is infinite so that I = V/R = 0, where V is the
potential difference between P and N Now,
V = Potential difference between P and A
+ Potential difference between A and B
+ Potential difference between B and N
= e
(3 37)
Thus, emf e is the potential difference between the positive and
negative electrodes in an open circuit, i e |
1 | 2881-2884 | Now,
V = Potential difference between P and A
+ Potential difference between A and B
+ Potential difference between B and N
= e
(3 37)
Thus, emf e is the potential difference between the positive and
negative electrodes in an open circuit, i e , when no current is
flowing through the cell |
1 | 2882-2885 | 37)
Thus, emf e is the potential difference between the positive and
negative electrodes in an open circuit, i e , when no current is
flowing through the cell If however R is finite, I is not zero |
1 | 2883-2886 | e , when no current is
flowing through the cell If however R is finite, I is not zero In that case the potential difference
between P and N is
V = V++ V– – I r
= e – I r
(3 |
1 | 2884-2887 | , when no current is
flowing through the cell If however R is finite, I is not zero In that case the potential difference
between P and N is
V = V++ V– – I r
= e – I r
(3 38)
Note the negative sign in the expression (I r) for the potential difference
between A and B |
1 | 2885-2888 | If however R is finite, I is not zero In that case the potential difference
between P and N is
V = V++ V– – I r
= e – I r
(3 38)
Note the negative sign in the expression (I r) for the potential difference
between A and B This is because the current I flows from B to A in the
electrolyte |
1 | 2886-2889 | In that case the potential difference
between P and N is
V = V++ V– – I r
= e – I r
(3 38)
Note the negative sign in the expression (I r) for the potential difference
between A and B This is because the current I flows from B to A in the
electrolyte In practical calculations, internal resistances of cells in the circuit
may be neglected when the current I is such that e >> I r |
1 | 2887-2890 | 38)
Note the negative sign in the expression (I r) for the potential difference
between A and B This is because the current I flows from B to A in the
electrolyte In practical calculations, internal resistances of cells in the circuit
may be neglected when the current I is such that e >> I r The actual
values of the internal resistances of cells vary from cell to cell |
1 | 2888-2891 | This is because the current I flows from B to A in the
electrolyte In practical calculations, internal resistances of cells in the circuit
may be neglected when the current I is such that e >> I r The actual
values of the internal resistances of cells vary from cell to cell The internal
resistance of dry cells, however, is much higher than the common
electrolytic cells |
1 | 2889-2892 | In practical calculations, internal resistances of cells in the circuit
may be neglected when the current I is such that e >> I r The actual
values of the internal resistances of cells vary from cell to cell The internal
resistance of dry cells, however, is much higher than the common
electrolytic cells We also observe that since V is the potential difference across R, we
have from Ohm’s law
V = I R
(3 |
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