Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5Ω is used in the external circuit of Read More …

Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R =10.0Ω  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 ?

Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R =10.0Ω  is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What Read More …

Figure 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 up to 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 its, 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.

Figure 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 up to Read More …

(a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance? 

(a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance? (b) Given the resistances of 1Ω, 2Ω,3Ω, how will Read More …

Choose the correct alternative: (a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.

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 Read More …

Answer the following questions: (a) A stead 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?

Answer the following questions: (a) A stead 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 Read More …

(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 380Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

(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 Read More …

The earth’s surface has a negative surface charge density of  10-9 Cm  . 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 neutralize the earth’s 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  6.37×106 m.)

The earth’s surface has a negative surface charge density of  10-9 Cm  . 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 Read More …

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