Resistance of the standard resistor, R = 10.0 Ω Balance point for this resistance, l₁ = 58.3 cm Current in the potentiometer wire = i Hence, potential drop across R, E₁ = iR Resistance of the unknown resistor = X Balance point for this resistor, l₂ = 68.5 cm Hence, potential drop across X, E2 = iX TRead more
Resistance of the standard resistor, R = 10.0 Ω
Balance point for this resistance, l₁ = 58.3 cm
Current in the potentiometer wire = i
Hence, potential drop across R, E₁ = iR
Resistance of the unknown resistor = X
Balance point for this resistor, l₂ = 68.5 cm
Hence, potential drop across X, E2 = iX
The relation connecting emf and balance point is,
E₁/E₂ =l₁/l₂
=> iR/iX =l₁/l₂
X= (l₁/l₂ ) x R
= 68.5/58.3 x 10 = 11.749 Ω
Therefore, the value of the unknown resistance, X, is 11.75Ω.
If we fail to find a balance point with the given cell of emf, ε , then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.
Ans (a). Constant emf of the given standard cell, E₁ = 1.02 V Balance point on the wire, l₁ = 67.3 cm A cell of unknown emf, ε , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm The relation connecting emf and balance point is, E₁/l₁ = ε /l ε = (l/l₁ ) x E₁ = (82.3 /Read more
Ans (a).
Constant emf of the given standard cell, E₁ = 1.02 V
Balance point on the wire, l₁ = 67.3 cm
A cell of unknown emf, ε , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm
The relation connecting emf and balance point is,
E₁/l₁ = ε /l
ε = (l/l₁ ) x E₁ = (82.3 /67.3 ) x 1.02 = 1.247 V
The value of unknown emf is 1.247 V.
Ans (b).
The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.
Ans (c).
The balance point is not affected by the presence of high resistance.
Ans (d).
The point is not affected by the internal resistance of the driver cell.
Ans (e).
The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V.This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.
Ans (f).
The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.
The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.
The resistance of each resistor connected in the given circuit, R = 1 Ω Equivalent resistance of the given circuit = R' The network is infinite. Hence, equivalent resistance is given by the relation, R' = 2 + R'/(R+1) (R')² -2 R' -2 = 0 R' = [2 ± √ (4+8)]/2 = [2 ± √12]/2 = 1 ± √3 Negative value of RRead more
The resistance of each resistor connected in the given circuit,
R = 1 Ω
Equivalent resistance of the given circuit = R’
The network is infinite.
Hence, equivalent resistance is given by the relation,
R’ = 2 + R’/(R+1)
(R’)² -2 R’ -2 = 0
R’ = [2 ± √ (4+8)]/2
= [2 ± √12]/2 = 1 ± √3
Negative value of R’ cannot be accepted.
Hence, equivalent resistance,
R’=(1 + √3)= 1 + 1.73 = 2.73 Ω
Internal resistance of the circuit, r = 0.5 Ω
Hence, total resistance of the given circuit
= 2.73 + 0.5 = 3.23 Ω
Supply voltage, V = 12 V
According to Ohm’s Law, current drawn from the source is given by the ratio, 12/3.23= 3.72 A
Ans (a). Alloys of metals usually have greater resistivity than that of their constituent metals. Ans (b). Alloys usually have lower temperature coefficients of resistance than pure metals. Ans (c). The resistivity of the alloy, manganin, is nearly independent of increase of temperature. Ans (d). ThRead more
Ans (a).
Alloys of metals usually have greater resistivity than that of their constituent metals.
Ans (b).
Alloys usually have lower temperature coefficients of resistance than pure metals.
Ans (c).
The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
Ans (d).
The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 1022.
Ans (a). When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant. Ans (b). No, ORead more
Ans (a).
When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant.
Ans (b).
No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.
Ans (c).
According to Ohm’s law, the relation for the potential is V = IR Voltage (V) is directly proportional to current (I).
R is the internal resistance of the source. I =V/R
If V is low, then R must be very low, so that high current can be drawn from the source.
Ans (d).
In order to prohibit the current from exceeding the safety limit, a high tension supply must have a very large internal resistance. If the internal resistance is not large, then the current drawn can exceed the safety limits in case of a short circuit.
Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε ?
Resistance of the standard resistor, R = 10.0 Ω Balance point for this resistance, l₁ = 58.3 cm Current in the potentiometer wire = i Hence, potential drop across R, E₁ = iR Resistance of the unknown resistor = X Balance point for this resistor, l₂ = 68.5 cm Hence, potential drop across X, E2 = iX TRead more
Resistance of the standard resistor, R = 10.0 Ω
Balance point for this resistance, l₁ = 58.3 cm
Current in the potentiometer wire = i
Hence, potential drop across R, E₁ = iR
Resistance of the unknown resistor = X
Balance point for this resistor, l₂ = 68.5 cm
Hence, potential drop across X, E2 = iX
The relation connecting emf and balance point is,
E₁/E₂ =l₁/l₂
=> iR/iX =l₁/l₂
X= (l₁/l₂ ) x R
= 68.5/58.3 x 10 = 11.749 Ω
Therefore, the value of the unknown resistance, X, is 11.75Ω.
If we fail to find a balance point with the given cell of emf, ε , then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.
See lessFigure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.(a) What is the value ε? (b) What purpose does the high resistance of 600 kΩ have?(c) Is the balance point affected by this high resistance? (d) Is the balance point affected by the internal resistance of the driver cell? (e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V? (f ) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
Ans (a). Constant emf of the given standard cell, E₁ = 1.02 V Balance point on the wire, l₁ = 67.3 cm A cell of unknown emf, ε , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm The relation connecting emf and balance point is, E₁/l₁ = ε /l ε = (l/l₁ ) x E₁ = (82.3 /Read more
Ans (a).
Constant emf of the given standard cell, E₁ = 1.02 V
Balance point on the wire, l₁ = 67.3 cm
A cell of unknown emf, ε , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm
The relation connecting emf and balance point is,
E₁/l₁ = ε /l
ε = (l/l₁ ) x E₁ = (82.3 /67.3 ) x 1.02 = 1.247 V
The value of unknown emf is 1.247 V.
Ans (b).
The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.
Ans (c).
The balance point is not affected by the presence of high resistance.
Ans (d).
The point is not affected by the internal resistance of the driver cell.
Ans (e).
The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V.This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.
Ans (f).
The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.
The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.
See lessDetermine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.
The resistance of each resistor connected in the given circuit, R = 1 Ω Equivalent resistance of the given circuit = R' The network is infinite. Hence, equivalent resistance is given by the relation, R' = 2 + R'/(R+1) (R')² -2 R' -2 = 0 R' = [2 ± √ (4+8)]/2 = [2 ± √12]/2 = 1 ± √3 Negative value of RRead more
The resistance of each resistor connected in the given circuit,
R = 1 Ω
Equivalent resistance of the given circuit = R’
The network is infinite.
Hence, equivalent resistance is given by the relation,
R’ = 2 + R’/(R+1)
(R’)² -2 R’ -2 = 0
R’ = [2 ± √ (4+8)]/2
= [2 ± √12]/2 = 1 ± √3
Negative value of R’ cannot be accepted.
Hence, equivalent resistance,
R’=(1 + √3)= 1 + 1.73 = 2.73 Ω
Internal resistance of the circuit, r = 0.5 Ω
Hence, total resistance of the given circuit
= 2.73 + 0.5 = 3.23 Ω
Supply voltage, V = 12 V
According to Ohm’s Law, current drawn from the source is given by the ratio, 12/3.23= 3.72 A
See lessChoose the correct alternative: (a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals. (b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals. (c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature. (d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10²²/10³ ).
Ans (a). Alloys of metals usually have greater resistivity than that of their constituent metals. Ans (b). Alloys usually have lower temperature coefficients of resistance than pure metals. Ans (c). The resistivity of the alloy, manganin, is nearly independent of increase of temperature. Ans (d). ThRead more
Ans (a).
Alloys of metals usually have greater resistivity than that of their constituent metals.
Ans (b).
Alloys usually have lower temperature coefficients of resistance than pure metals.
Ans (c).
The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
Ans (d).
See lessThe resistivity of a typical insulator is greater than that of a metal by a factor of the order of 1022.
Answer the following questions: (a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed? (b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law. (c) A low voltage supply from which one needs high currents must have very low internal resistance. Why? (d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?
Ans (a). When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant. Ans (b). No, ORead more
Ans (a).
When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant.
Ans (b).
No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.
Ans (c).
According to Ohm’s law, the relation for the potential is V = IR Voltage (V) is directly proportional to current (I).
R is the internal resistance of the source. I =V/R
If V is low, then R must be very low, so that high current can be drawn from the source.
Ans (d).
In order to prohibit the current from exceeding the safety limit, a high tension supply must have a very large internal resistance. If the internal resistance is not large, then the current drawn can exceed the safety limits in case of a short circuit.
See less