3.21 ADDITIONAL EXERCISES

Q3.14 The earth’s surface has a negative surface charge density of 10^−9 C m^−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 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 × 10^6 m.)

A3.14 Take the radius of the earth = 6.37 × 10^6 m and obtain total charge of the globe. Divide it by current to obtain time = 283 s. Still this method gives you only an estimate; it is not strictly correct. Why?

Q3.15


(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?

A3.15


(a) 1.4 A, 11.9 V


(b) 0.005 A; impossible because a starter motor requires large current (~ 100 A) for a few seconds.

Q3.16 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^−8 Ωm, ρCu = 1.72 × 10^−8 Ωm, Relative density of Al = 2.7, of Cu = 8.9.)

A3.16 The mass (or weight) ratio of copper to aluminium wire is (1.72/2.63) × (8.9/2.7) ≅ 2.2. Since aluminium is lighter, it is preferred for long suspensions of cables.

Q3.17 What conclusion can you draw from the following observations on a resistor made of alloy manganin?

A3.17 Ohm’s law is valid to a high accuracy; the resistivity of the alloy manganin is nearly independent of temperature.

Q3.18 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?

A3.18


(a) Only current (because it is given to be steady!). The rest depends on the area of cross-section inversely.


(b) No, examples of non-ohmic elements: vacuum diode, semiconductor diode.


(c) Because the maximum current drawn from a source = ε/r


(d) Because, if the circuit is shorted (accidentally), the current drawn will exceed safety limits, if internal resistance is not large.

Q3.19 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 (10^22/10^23).

A3.19


(a) greater,


(b) lower,


(c) nearly independent of,


(d) 10^22.

Q3.20


(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 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?


(c) Determine the equivalent resistance of networks shown in Fig. 3.31.

Fig. 3.31.

A3.20


(a) (i) in series, (ii) all in parallel; n^2.


(b) (i) Join 1 Ω, 2 Ω in parallel and the combination in series with 3 Ω, (ii) parallel combination of 2 Ω and 3 Ω in series with 1 Ω, (iii) all in series, (iv) all in parallel.


(c) (i) (16/3) Ω, (ii) 5 R.

Q3.21 Determine the current drawn from a 12V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1 Ω resistance.

Fig. 3.32.

A3.21 Hint: Let X be the equivalent resistance of the infinite network. Clearly, 2 + X/(X +1) = X which gives X = (1 + √3 ) Ω; therefore the current is 3.7 A.

Q3.22 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 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.

Figure 3.33

(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) 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?


(e) 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?

A3.22


(a) ε = 1.25 V.


(b) To reduce current through the galvanometer when the movable contact is far from the balance point.


(c) No.


(d) No.


(e) No. If ε is greater than the EMF of the driver cell of the potentiometer, there will be no balance point on the wire AB.


(f) The circuit, as it is, would be unsuitable, because the balance point (for ε of the order of a few mV) will be very close to the end A and the percentage error in measurement will be very large. The circuit is modified by putting a suitable resistor R in series with the wire AB so that potential drop across AB is only slightly greater than the EMF to be measured. Then, the balance point will be at larger length of the wire and the percentage error will be much smaller.

Q3.23 Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω 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 3.34

A3.23 1.7 Ω

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End of Chapter 3 Current Electricity for blind and visually impaired students.