It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm's law. According to Ohm's law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistRead more
It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7 Ω .
Resistivity of aluminium, ρAl = 2.63 x 10-8 Ω m Relative density of aluminium, d₁= 2.7 Let l₁ be the length of aluminium wire and m₁ be its mass. Resistance of the aluminium wire = R₁ Area of cross-section of the aluminium wire = A₁ Resistivity of copper, ρcu = 1.72 x 10-8 Ωm Relative density of copRead more
Resistivity of aluminium, ρAl = 2.63 x 10-8 Ω m
Relative density of aluminium, d₁= 2.7
Let l₁ be the length of aluminium wire and m₁ be its mass.
Resistance of the aluminium wire = R₁
Area of cross-section of the aluminium wire = A₁
Resistivity of copper, ρcu = 1.72 x 10-8 Ωm
Relative density of copper, d₂ = 8.9
Let l₂ be the length of copper wire and m₂ be its mass.
Resistance of the copper wire = R2
Area of cross-section of the copper wire = A2
The two relations can be written as
R₁ =ρ₁ l₁/A₁ ——————–Eq-1
R2 =ρ2 l2A2 ——————–Eq-2
It is given that ,R₁=R2 => ρ₁ l₁/A₁= ρ2 l2A2
=> ρ₁ /A₁ = ρ2 /A2 ( as ,l₁=l2)
=> A₁ /A2 =ρ₁/ρ2 = (2.63 x 10⁻⁸ )/(1.72 x 10⁻⁸ )=2.63/1.72
Mass of the copper wire ,m₁= Volume x Density
=A₁L₁ x d₁ =A₁l₁d₁———————-Eq -3
Mass of the copper wire ,m2= Volume x Density
=A2L2 x d2 =A2l2d2———————-Eq -4,from which we obtain
m₁/m2 = (A1l1d1)/(A2l2d2)
For l1=l2
m1/m2 = A1d1/A2d2
For A1/A2 =2.63/1.72,
Therefore
m1/m2= 2.63/1.72 x 2.7/8.9 = 0.46
It can be inferred from this ratio that m1 is less than m2. Hence, aluminium is lighter than copper. Since aluminium is lighter, it is preferred for overhead power cables over copper.
Ans (a). Number of secondary cells, n = 6 Emf of each secondary cell, E = 2.0 V Internal resistance of each cell, r = 0.015 Ω series resistor is connected to the combination of cells. Resistance of the resistor, R = 8.5 Ω Current drawn from the supply = I, which is given by the relation, I = (nE)/(RRead more
Ans (a).
Number of secondary cells, n = 6
Emf of each secondary cell, E = 2.0 V
Internal resistance of each cell, r = 0.015 Ω series resistor is connected to the combination of cells. Resistance of the resistor, R = 8.5 Ω
Current drawn from the supply = I, which is given by the relation,
I = (nE)/(R+nr)
= (6×2)/(8.5 + 6 x 0.015) = 12 /8.59 =1.39 A
Terminal voltage, V = IR = 1.39 x 8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.
Ans (b).
After a long use, emf of the secondary cell, E = 1.9 V
Internal resistance of the cell, r = 380 Ω
Hence, maximum current =E/r=1.9/380 = 0.005 A
Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.
Number density of free electrons in a copper conductor, n = 8.5 x 1028 m-3 Length of the copper wire, I = 3.0 m Area of cross-section of the wire, A = 2.0 x 10-6 m2 Current carried by the wire, I = 3.0 A, which is given by the relation, I = nAeVd Where, e = Electric charge = 1.6 x 10-19 C Vd= DriftRead more
Number density of free electrons in a copper conductor, n = 8.5 x 1028 m–3 Length of the copper wire, I = 3.0 m
Area of cross-section of the wire, A = 2.0 x 10-6 m2
Current carried by the wire, I = 3.0 A, which is given by the relation, I = nAeVd Where, e = Electric charge = 1.6 x 10-19 C
Vd= Drift Velocity = Length of the wire (l) /Time taken to cover l (t)
I = nAel/t => t = nAel/t
t = 3x 8.5 x 1028 x 2 x 10-6 x 1.6 x 10-19 )/3.0
= 2.7 x 10⁴ s
Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 x 104 s.
Emf of the cell, E₁ = 1.25 V Balance point of the potentiometer, l₁= 35 cm. The cell is replaced by another cell of emf E2. New balance point of the potentiometer, l2 = 63 cm The balance condition is given by the relation. E₁/E2= l₁/l2 =>E2 = = E₁l2/l₁ = 1.25 x 63/35 =2.25 V Therefore, emf ofRead more
Emf of the cell, E₁ = 1.25 V
Balance point of the potentiometer, l₁= 35 cm. The cell is replaced by another cell of emf E2.
New balance point of the potentiometer, l2 = 63 cm The balance condition is given by the relation.
What conclusion can you draw from the following observations on a resistor made of alloy manganin?
It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm's law. According to Ohm's law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistRead more
It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7 Ω .
See lessTwo wires of equal length, one of aluminum and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminum wires are preferred for overhead power cables. (ρAl = 2.63 × 10⁻⁸ Ω m, ρCu = 1.72 × 10⁻⁸Ω m, Relative density of Al = 2.7, of Cu = 8.9.)
Resistivity of aluminium, ρAl = 2.63 x 10-8 Ω m Relative density of aluminium, d₁= 2.7 Let l₁ be the length of aluminium wire and m₁ be its mass. Resistance of the aluminium wire = R₁ Area of cross-section of the aluminium wire = A₁ Resistivity of copper, ρcu = 1.72 x 10-8 Ωm Relative density of copRead more
Resistivity of aluminium, ρAl = 2.63 x 10-8 Ω m
Relative density of aluminium, d₁= 2.7
Let l₁ be the length of aluminium wire and m₁ be its mass.
Resistance of the aluminium wire = R₁
Area of cross-section of the aluminium wire = A₁
Resistivity of copper, ρcu = 1.72 x 10-8 Ωm
Relative density of copper, d₂ = 8.9
Let l₂ be the length of copper wire and m₂ be its mass.
Resistance of the copper wire = R2
Area of cross-section of the copper wire = A2
The two relations can be written as
R₁ =ρ₁ l₁/A₁ ——————–Eq-1
R2 =ρ2 l2A2 ——————–Eq-2
It is given that ,R₁=R2 => ρ₁ l₁/A₁= ρ2 l2A2
=> ρ₁ /A₁ = ρ2 /A2 ( as ,l₁=l2)
=> A₁ /A2 =ρ₁/ρ2 = (2.63 x 10⁻⁸ )/(1.72 x 10⁻⁸ )=2.63/1.72
Mass of the copper wire ,m₁= Volume x Density
=A₁L₁ x d₁ =A₁l₁d₁———————-Eq -3
Mass of the copper wire ,m2= Volume x Density
=A2L2 x d2 =A2l2d2———————-Eq -4,from which we obtain
m₁/m2 = (A1l1d1)/(A2l2d2)
For l1=l2
m1/m2 = A1d1/A2d2
For A1/A2 =2.63/1.72,
Therefore
m1/m2= 2.63/1.72 x 2.7/8.9 = 0.46
It can be inferred from this ratio that m1 is less than m2. Hence, aluminium is lighter than copper. Since aluminium is lighter, it is preferred for overhead power cables over copper.
See less(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage? (b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?
Ans (a). Number of secondary cells, n = 6 Emf of each secondary cell, E = 2.0 V Internal resistance of each cell, r = 0.015 Ω series resistor is connected to the combination of cells. Resistance of the resistor, R = 8.5 Ω Current drawn from the supply = I, which is given by the relation, I = (nE)/(RRead more
Ans (a).
Number of secondary cells, n = 6
Emf of each secondary cell, E = 2.0 V
Internal resistance of each cell, r = 0.015 Ω series resistor is connected to the combination of cells. Resistance of the resistor, R = 8.5 Ω
Current drawn from the supply = I, which is given by the relation,
I = (nE)/(R+nr)
= (6×2)/(8.5 + 6 x 0.015) = 12 /8.59 =1.39 A
Terminal voltage, V = IR = 1.39 x 8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.
Ans (b).
After a long use, emf of the secondary cell, E = 1.9 V
Internal resistance of the cell, r = 380 Ω
Hence, maximum current =E/r=1.9/380 = 0.005 A
Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.
See lessThe number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10²⁸ m⁻³. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10⁻⁶ m² and it is carrying a current of 3.0 A.
Number density of free electrons in a copper conductor, n = 8.5 x 1028 m-3 Length of the copper wire, I = 3.0 m Area of cross-section of the wire, A = 2.0 x 10-6 m2 Current carried by the wire, I = 3.0 A, which is given by the relation, I = nAeVd Where, e = Electric charge = 1.6 x 10-19 C Vd= DriftRead more
Number density of free electrons in a copper conductor, n = 8.5 x 1028 m–3 Length of the copper wire, I = 3.0 m
Area of cross-section of the wire, A = 2.0 x 10-6 m2
Current carried by the wire, I = 3.0 A, which is given by the relation, I = nAeVd Where, e = Electric charge = 1.6 x 10-19 C
Vd= Drift Velocity = Length of the wire (l) /Time taken to cover l (t)
I = nAel/t => t = nAel/t
t = 3x 8.5 x 1028 x 2 x 10-6 x 1.6 x 10-19 )/3.0
= 2.7 x 10⁴ s
Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 x 104 s.
See lessIn a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Emf of the cell, E₁ = 1.25 V Balance point of the potentiometer, l₁= 35 cm. The cell is replaced by another cell of emf E2. New balance point of the potentiometer, l2 = 63 cm The balance condition is given by the relation. E₁/E2= l₁/l2 =>E2 = = E₁l2/l₁ = 1.25 x 63/35 =2.25 V Therefore, emf ofRead more
Emf of the cell, E₁ = 1.25 V
Balance point of the potentiometer, l₁= 35 cm. The cell is replaced by another cell of emf E2.
New balance point of the potentiometer, l2 = 63 cm The balance condition is given by the relation.
E₁/E2= l₁/l2
=>E2 =
= E₁l2/l₁ = 1.25 x 63/35 =2.25 V
Therefore, emf of the second cell is 2.25V.
See less