Some of the most important Physics Topics include energy, motion, and force.
Define Mobility of Electrons ? What is the Formula for Mobility of Electrons?
Equilibrium of drift velocity: Suppose a potential difference of V is applied at the two ends of a homogeneous conductor of length l. The magnitude of uniform electric field produced inside the conductor is,
E = \(\frac{V}{l}\) …. (1)
So, the force acting on a free electron inside the conductor,
eE = \(\frac{e V}{l}\) …. (2)
Acceleration of a free electron = \(\frac{\mathrm{eV}}{\mathrm{ml}}\); (where m is the mass of the electron = 9.1 × 10-31 kg)
Due to this acceleration the velocity of the electron will continu-ously increase. But practically it does not happen. The motion of the electron is thwarted by its collision with the atoms and the ions inside the conductor. If this opposing force is considered to be equivalent to viscous force then the force acting against the motion of the electron inside the metallic conductor is pro-portional to the velocity of the electron i.e., opposing force = kv [v = velocity of the electron; k = constant].
This opposing force increases with the increase of the velocity of the electron. At one time this force will be equal to the force due to the electric field. Since these two forces are opposite to each other the resultant force becomes zero. Then the electron has no acceleration and it moves with uniform velocity inside the metallic conductor. This uniform velocity of the electron is the drift velocity v, of the free electrons. So the condition of equilib-rium is
eE = kvd
or vd = \(\frac{e E}{k}\) = µE …… (3)
This constant µ ( = \(\frac{e}{k}\) or \(\frac{v_d}{E}\)) is called the mobility of the free electrons. Clearly, if E = 1 then µ = vd.
Definition: Mobility of a free electron is the uniform drift velocity attained by it due to the application of unit uniform electric field inside the metallic conductor.
Unit of µ : As µ = \(\frac{v_d}{E}\)
So, unit of µ \(=\frac{\text { unit of } v_d}{\text { unit of } E}\) = \(\frac{\mathrm{m} \cdot \mathrm{s}^{-1}}{\mathrm{~V} \cdot \mathrm{m}^{-1}}\) = m2 ᐧ V-1 ᐧ s-1
We have, vd = \(\frac{I}{n e A}\)
But according to equations (1) and (3) of section 1.15.2,
vd = \(\frac{e E}{k}\) = \(\frac{e V}{k l}\)
so, \(\frac{e V}{k l}\) = \(\frac{I}{n e A}\)
or, V = \(\frac{k}{n e^2} \cdot \frac{l}{A} \cdot I\) ….. (1)
For a fixed conductor l and A are constants. So V ∝ I; this is Ohm’s law.
If ρ be the resistivity of a metal, its resistance R = ρ\(\frac{l}{A}\).
So, V = RI = ρ\(\frac{l}{A}\) ᐧ I.
Comparing equations (1) and (2) we get,
resistivity, ρ = \(\frac{k}{n e^2}\)
i.e., conductivity of the metal, σ = \(\frac{1}{\rho}\) = \(\frac{n e^2}{k}\)
So, conductivity of a metal depends on the number density of free electrons. Since, the value of n differs from metal to metal, conductivity (i.e., resistivity) also differs from metal to metal.
Vector form of Ohms law: Let us consider a conductor of length = l, cross sectional area = A, resistivity of the material = ρ.
Then the resistance of the conductor,
R = ρ ᐧ \(\frac{l}{A}\) = \(\frac{l}{\sigma A}\) [σ = conductivity = \(\frac{1}{\rho}\)]
If the potential difference between the two ends of the conductor be V, then the electric field generated in it,
E = \(\frac{V}{l}\) i.e., V = El
On the other hand, if the current in the conductor be I, then current density, j = \(\frac{I}{A}\) i.e., I = jA
Following Ohm’s law,
V = IR or, El = jA ᐧ \(\frac{l}{\sigma A}\)
or, E = \(\frac{1}{\sigma} \cdot j\) or, j = σE
Here j and E are vectors.
So, \(\vec{j}\) = σ\(\vec{E}\) – this equation is the vector form of Ohm’s law.
Numerical Examples
Example 1.
A 100 V battery has an internal resistance 3 Ω. What is the reading of a voltmeter having resistance 200Ω, when placed across the terminals of the battery? What should be the minimum value of the voltmeter resis-tance so that the error in finding the emf of the battery may not be more than 1 % ? [HS 02]
Solution:
Main current, I = \(\frac{E}{R+r}\) = \(\frac{100}{200+3}\) = \(\frac{100}{203} \mathrm{~A}\)
So, lost volt = Ir = \(\frac{100}{203}\) × 3 = \(\frac{300}{203}\) = 1.48 V
Therefore, reading of the voltmeter,
V = E – Ir = 100 – 1.48 = 98.52 V
In the second case, error = 1% = \(\frac{1}{100}\)
Reading of the voltmeter,
V = E – \(\frac{E}{100}\) = 100 – 1 = 99 V
So, lost volt = 100 – 99 = 1 V
i.e., Ir = 1 or, I = \(\frac{\mathbf{l}}{r}\) = \(\frac{1}{3} \mathrm{~A}\)
This current passes through the voltmeter. So resistance of the voltameter = \(\frac{V}{I}\) = \(\frac{99}{\frac{1}{3}}\) = 297 Ω
Example 2.
A wire of resistance 10 Ω is used to form a circular ring of circumference 10 cm. If two current carrying conductors are connected at any two points, the sub circuit so formed has a resistance of 1 Ω. Find the positions of the two points.
Solution:
Current carrying conductors are connected at the two points A and B of the circular ring [Fig.]. According to the question,
R1 + R2 = 10 Ω ……… (1)
The two parts having resistances R1 and R2 form a parallel combination at A and B whose equivalent resistance is 1 Ω.
Since resistance of the wire of length 10 cm is 10 Ω, the resis-tance of the wire of length 1 cm is 1 Ω
Therefore, the lengths of the two portions of the wire having resistances R1 and R2 are 8.873 cm and 1.127 cm.
Example 3.
12 cells each having the same emf are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery is connected in series with an ammeter and two other similar cells. The current is 3 A when the two cells aid the battery and is 2 A when the cells and the battery oppose each other. How many cells in the battery are wrongly con-nected?
Solution:
If n -number of cells are wrongly connected, then the number of cells correctly connected in the battery = 12 – n; emf of each cell = E.
∴ Emf of the battery = (12 – n)E – nE = (12 – 2n)E
Again, emf of the additional two cells = 2E. Let R be the resis-tance of the whole circuit.
In the first case,
So, only one cell was wrongly connected.
Example 4.
A cell of emf 1.4 V and internal resistance 2 Ω is con-nected in series with a resistance of 100 Ω and an ammeter. The resistance of the ammeter is \(\frac{4}{3}\) Ω. To measure the potential difference between the two ends of the resistance a voltmeter is connected,
(i) Draw the circuit
(ii) If the reading of the ammeter is 0.02 A, what is the resistance of the voltmeter?
(iii) If the reading of the voltmeter is 1.10 V, what will be its error?
Solution:
i) Fig. is the circuit diagram.
ii) Lost volt of the cell = Ir = 0.02 × 2 = 0.04 V
Potential difference between the two ends of the ammeter
iii) Voltmeter should record VCD = \(\frac{4}{3}\) = 1.33 V
∴ Error in the reading of the voltmeter
= 1.33 – 1.10 = 0.23V
Example 5.
In the given Fig., what is the equivalent resistance between the two points A and B’
Solution:
The two points A and C are connected by a connecting wires having no resistance. So their potentials are equal. i.e. the two points A and C are identical. Similarly, the two points B and D are identical. So the circuit shown in the Fig. is the equivalent circuit of the Fig.
Therefore, if r be the equivalent resistance between the two points A and B, then
Example 6.
A circuit is given in the Fig. The emf of the battery is 1.8 V and internal resistance is \(\frac{2}{3}\) Ω. Calculate the current through the 3 Ω resistance. What is the amount of dissipated energy in the whole circuit?
Solution:
The potentials of the two points C and D are equal to that of the point A. Again, the potentials of the two points C’ and D’ are equal to that of the point B. So the circuit shown in Fig. is the equivalent circuit of the Fig.
In Fig. the resistance of the middle branch between A and B is
The amount of dissipated energy in the whole circuit = emf of the battery × main current
[See the chapter ‘Electrical Energy and Power’)
= 1.8 × 0.9 = 1.62W
Example 7.
An infinite ladder network of resistances is constructed with 1 Ω and 2 Ω resistances as shown in Fig. The 6V battery between A and B has negligible internal resistance.
(i) Show that the effective resistance between A and B is 2 Ω.
(ii) What is the current that passes through the 2 Ω resistance nearest to the battery?
Solution:
Suppose, effective resistance between A and B = R. From the Fig. it is clear that on the right side of CD the infinite network is equivalent to that on the right side of AB because deleting one chain does not affect the equivalent resistance R. So, the circuit shown in the Fig. is the equivalent circuit of the Fig.
i) From the Fig., effective resistance between A and B is
R = 1 + \(\frac{2 R}{2+R}\) or, R = \(\frac{2+3 R}{2+R}\)
or, R2 – R – 2 = 0 or, (R + 1)(R – 2) = 0
Obviously, R ≠ -1
Hence, R = 2 Ω
ii) Main current, I = \(\frac{6}{R}\) = \(\frac{6}{2}\) = 3 A
∴ VAC = I × 1 = 3 × 1 = 3V
∴ VCD = 6 – 3 = 3V
So, current through the 2 Ω resistance nearest to the battery
= \(\frac{V_{C D}}{2}\) = \(\frac{3}{2}\) = 1.5 A
Example 8.
In the circuit shown in the Fig., calculate the direct current (dc) through the 2 Ω resistance. The Internal resistance of the battery is negligible and C = 0.2µF.
Solution:
Direct current cannot pass through any capacitor. So no current flows through the 4 Ω resistance.
Now, equivalent resistance between A and B
= \(\frac{2 \times 3}{2+3}\) = \(\frac{6}{5}\) = 1.2 A
So, main current, I = \(\frac{6}{1.2+2.8}\) = \(\frac{6}{4}\) = 1.5 A
Since no current passes through a capacitor in the steady state, the branch containing C and 4 Ω has been treated as deleted.
∴ VAB = I × 1.2 = 1.5 × 1.2 = 1.8 V
∴ Current through 2 Ω resistance = \(\frac{V_{A B}}{2}\) = \(\frac{1.8}{2}\) = 0.9 A
Example 9.
Three resistances A, B and C are connected in such a way that their combined equivalent resistance is equal to that of B. If A and B are 10 Ω and 30 Ω respectively, find the three possible values of C and draw the corresponding circuits.
Solution:
In this case there are only three possible arrangements which are shown in the Fig. If the three resistances are arranged in any other alternative way the equivalent resistance will he greater than B or less than B.
In each arrangement, the equivalent resistance is equal to B (given).
So, in the arrangement (a),
A + \(\frac{B C}{B+C}\) = B, 10 + \(\frac{30 C}{30+C}\) = 30
or, 30C = 600 + 20C or, bC 600 or, C = 60 Ω
In the arrangement (b),
\(\frac{A B}{A+B}\) + C = B or, \(\frac{10 \times 30}{10+30}\) + C = 30
or, C = 30 – \(\frac{30}{4}\) = 30 – 7.5 = 22.5 Ω
In the arrangement (c),
\(\frac{(A+B) C}{(A+B)+C}\) = B or, \(\frac{(10+30) \times C}{(10+30)+C}\) = 30
or, \(\frac{40 C}{40+C}\) = 30, 40C = 1200 + 30C
or 10C = 1200 or, C = 120 Ω
Example 10.
Two cells, one of emf 1.4 V and internal resistance 0.6 Ω, the other of emf 2.5 V and internal resistance 0.3 Ω are connected in parallel and the combination is connected in series with an external resistance of 4 Ω. What is the current through this resistance?
Solution:
Suppose in Fig., the potential difference between A and B, VA – VB = V.