Question:

What conclusion can you draw from the following observations on a resistor made of alloy manganin? Current A Voltage V Current A Voltage V 0.2 3.94 3.0 59.2 0.4 7.87 4.0 78.8 0.6 11.8 5.0 98.6 0.8 15.7 6.0 118.5 1.0 19.7 7.0 138.2 2.0 39.4 8.0 158.0

Answer:

Step 1: Observe the data provided.

Step 2: Calculate the voltage for each current value.

Step 3: Plot the data points on a graph.

Step 4: Draw a line of best fit through the data points.

Step 5: Calculate the slope of the line of best fit.

Step 6: The conclusion is that the resistor made of alloy manganin has a linear relationship between current and voltage, with a slope of 19.7 volts per ampere.

Question:

A battery of emf 10 V and internal resistance 3 W is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer:

1. Resistance of the resistor = 10 V / 0.5 A = 20 Ω
2. Terminal voltage of the battery = 10 V - (0.5 A x 3 Ω) = 8.5 V

Question:

(a) Three resistors 1Ω,2Ω and 3Ω are combined in series. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Answer:

(a) The total resistance of the combination is 6Ω.

(b) The potential drop across each resistor can be calculated using Ohm’s Law:

Resistor 1: V = I x R = 12V/6Ω = 2V

Resistor 2: V = I x R = 12V/6Ω = 2V

Resistor 3: V = I x R = 12V/6Ω = 2V

Question:

A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0oC? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70×10^−4oC^−1.

Answer:

1. Calculate the resistance of the heating element using Ohm’s Law: R = V/I = 230/3.2 = 71.875 Ω

2. Calculate the temperature of the heating element using the formula: T = (R2 - R1)/(α*R1) + T1 where R2 is the resistance of the heating element at the steady temperature, R1 is the initial resistance of the heating element, α is the temperature coefficient of resistance of nichrome, and T1 is the initial temperature of the heating element.

3. Substitute the values in the formula: T = (71.875 - 71.875)/(1.7 x 10^-4 x 71.875) + 27.0 = 27.0oC

4. Therefore, the steady temperature of the heating element is 27.0oC.

Question:

Choose 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).

Answer:

(a) Greater (b) Lower (c) Nearly independent of (d) 1022

Question:

The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Answer:

1. Determine the maximum current that can be drawn from the battery:

Maximum current = emf/resistance = 12 V/0.4 Ω = 30 A

Question:

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0×10^−7m^2, and its resistance is measured to be 5.0Ω. What is the resistivity of the material at the temperature of the experiment?

Answer:

Step 1: Calculate the area of the wire. A = 6.0 × 10^−7 m^2

Step 2: Calculate the resistance of the wire. R = 5.0 Ω

Step 3: Calculate the resistivity of the material. ρ = R × A ρ = 5.0 Ω × 6.0 × 10^−7 m^2 ρ = 3.0 × 10^−7 Ωm

Question:

Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρAl​=2.63×10^−8Ωm,ρCu​=1.72×10^−8Ωm, Relative density of Al = 2.7, of Cu = 8.9.)

Answer:

Step 1: Since the two wires have the same resistance, the wire that is lighter is the aluminium wire.

Step 2: This is because the resistivity of aluminium is lower than that of copper (2.63×10^−8Ωm compared to 1.72×10^−8Ωm). This means that the same amount of current will flow through a given length of aluminium wire as through a given length of copper wire, but the aluminium wire will be lighter.

Step 3: Aluminium wires are preferred for overhead power cables because they are lighter than copper wires and can therefore be used to make longer cables. Additionally, they are less expensive than copper wires and have a higher relative density, making them more durable and resistant to corrosion.

Question:

Answer the following question: (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 Ohms law universally applicable for all conducting elements?If not, give examples of elements which do not obey Ohms 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?

Answer:

Answer (a) Current is the only quantity that is constant along the conductor.

Answer (b) No, Ohm’s law is not universally applicable for all conducting elements. Examples of elements which do not obey Ohm’s law include superconductors, semiconductors, and non-ohmic materials.

Answer (c) A low voltage supply from which one needs high currents must have very low internal resistance because a higher internal resistance would result in a voltage drop across the supply, reducing the amount of current that can be drawn from it.

Answer (d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance because a lower internal resistance would result in a high current flow, which could lead to a short circuit and potentially cause a fire.

Question:

(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015Ω are joined in series to provide a supply to a resistance of 8.5Ω . 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 380Ω . What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Answer:

(a) Current drawn from the supply = 6 x 2.0 V / (8.5Ω + 0.015Ω) = 0.7 A

Terminal voltage = 6 x 2.0 V = 12 V

(b) Maximum current that can be drawn from the cell = 1.9 V / 380Ω = 0.005 A

No, the cell cannot drive the starting motor of a car as the current is too low.

Question:

A silver wire has a resistance of 2.1Ω at 27.5oC, and a resistance of 2.7Ω at 100oC. Determine the temperature coefficient of resistivity of silver.

Answer:

1. Find the change in resistance: Change in resistance = 2.7Ω - 2.1Ω = 0.6Ω

2. Find the change in temperature: Change in temperature = 100oC - 27.5oC = 72.5oC

3. Calculate the temperature coefficient of resistivity: Temperature coefficient of resistivity = Change in resistance/Change in temperature = 0.6Ω/72.5oC = 0.00826Ω/oC

Question:

A storage battery of emf 8.0 V and internal resistance 0.5Ω is being charged by a 120 V dc supply using a series resistor of 15.5Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Answer:

Answer:

1. The terminal voltage of the battery during charging is 105.5 V.
2. The purpose of having a series resistor in the charging circuit is to reduce the current flowing through the battery and to protect the battery from being overcharged.

Question:

(a) Three resistors 2Ω,4Ω and 5Ω are combined in parallel. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Answer:

(a) Total resistance = 1/R = 1/2 + 1/4 + 1/5 = 11/20 R = 20/11 Ω

(b) Total current drawn from the battery, I = 20/20/11 = 2/11 A

Current through 2Ω resistor, I2 = 2/20/11 = 1/11 A

Current through 4Ω resistor, I4 = 4/20/11 = 2/11 A

Current through 5Ω resistor, I5 = 5/20/11 = 5/11 A

Question:

The earths surface has a negative surface charge density of 10^−9Cm^−2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earths surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37×10^6 m.)

Answer:

1. Calculate the charge on the earth’s surface: Charge = surface charge density x area of earth’s surface Charge = 10^−9 Cm^−2 x 4πr^2 Charge = 8.08 x 10^14 C

2. Calculate the current required to neutralise the charge: Current = charge/time Current = 8.08 x 10^14 C/t

3. Calculate the time required to neutralise the charge: Time = charge/current Time = 8.08 x 10^14 C/1800 A Time = 4.49 x 10^12 s Time = 14.5 million years

Question:

At room temperature (27.0oC) the resistance of a heating element is 100Ω . What is the temperature of the element if the resistance is found to be 117Ω, given that the temperature coefficient of the material of the resistor is 1.70×10^−4C^−1.

Answer:

1. Calculate the temperature coefficient of the resistor: Temperature coefficient = 1.70 x 10^-4 C^-1

2. Calculate the change in temperature: Change in temperature = (117Ω - 100Ω) / (1.70 x 10^-4 C^-1) Change in temperature = 17 / (1.70 x 10^-4) Change in temperature = 10,000 C

3. Calculate the new temperature of the element: New temperature = 27.0oC + 10,000 C New temperature = 10,000.0oC

Question:

In 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?

Answer:

1. Calculate the difference between the balance points of the two cells: 63.0 cm - 35.0 cm = 28.0 cm

2. Calculate the ratio of the two cells’ emfs: 28.0 cm/1.25 V = 22.4

3. Calculate the emf of the second cell: 22.4 x 1.25 V = 28.0 V