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Chapter 3 Current Electricity Solutions

Question - 11 : - A storage battery of emf 8.0 V and internalresistance 0.5 Ω is being charged by a 120 V dc supply using a series resistorof 15.5 Ω. What is the terminal voltage of the battery during charging? What isthe purpose of having a series resistor in the charging circuit?

Answer - 11 : -

Now, terminal voltage of the battery duringcharging
V = E + lr = 8 + 7(0.5) = 11.5 V A series resistance is joined in the chargingcircuit to limit the excessive current so that charging is slow and permanent.

Question - 12 : - In a potentiometer arrangement, a cell of emf1.25 V gives a balance point at 35.0 cm length of the wire. If cell is replacedby another cell and the balance point shifts to 63.0 cm, what is the emf of thesecond cell?

Answer - 12 : - The potential gradient remains the same, as there is no change in thesetting of standard circuit.

Question - 13 : - The number density of free electrons in acopper conductor estimated is ,8.5 × 1028 m-3. Howlong does an electron take to drift from one end of a wire 3.0 m long to itsother end? The area of cross-section of the wire is 2.0 × 10-6 m2 andit is carrying a current of 3.0 A.

Answer - 13 : - We can first calculate drift velocity of the electrons from the givendata I = Aneud

Question - 14 : - Theearth’s surface has a negative surfacecharge density of 10-9  cm-2. The potentialdifference 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 1800A over the entire globe. If there were no mechanism of sustaining atmosphericelectric field, how much time (roughly) would be required to neutralise theearth’s surface? (This never happens in practice because there is a mechanismto replenish electric charges, namely the continual thunderstorms and lightningin different parts of the globe). (Radius of earth = 6.37 × 106 m).

Answer - 14 : - Due to negative charge on earth, an electric field is into the earthsurface due to which the positive ions of atmosphere are constantly pumped inand an equivalent current of 1800 A is established across the globe. Let usfirst calculate total negative charge on earth

Question - 15 : -

(a) Sixlead-acid type of secondary cells each of emf 2.0 V and internal resistance of0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. Whatare the current drawn from the supply and its terminal voltage?

(b) Asecondary cell after long use has an emf of 1.9 V and a large internalresistance of 380 Ω. What maximum current can be drawn from the cell ? Couldthe cell drive the starting motor of a car ?

Answer - 15 : - (a) Six cells are joined in series.

Equivalent emf is 2 × 6 = 12 V
Equivalent internal resistance is 0.015 × 6 = 0.09 Ω
Current drawn from supply
(b) Maximum current is drawn from a battery when external resistanceis tested to be zero

To start a car, a current of the order of 100A is needed, so the battery mentioned above can not drive the starting motor.

Question - 16 : - Two wires of equal length, one of aluminiumand the other of copper have the same resistance. Which of the two wires islighter? Hence explain why aluminium wires are preferred for overhead powercables.

Answer - 16 : - Two wires have same length, and resistance. As the specific resistancesare unequal, the areas are different. For copper wire, Rcu = pcu 

Thus, the aluminium wire for the sameresistance is very light than copper and that is why aluminium wires arepreferred for overhead power cables.

Question - 17 : - What conclusion can you draw from thefollowing observations on a resistor made of alloy manganin?

Answer - 17 : - We can find resistance of the alloy manganin for all the readings asfollows :

Here we can conclude that Ohm’s law is validto a high accuracy and resistance of the alloy is nearly constant at allcurrents.

Question - 18 : -

Answer the followingquestions:

  1. 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?
  2. Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.
  3. A low voltage supply from which one needs high currents must have very low internal resistance. Why?
  4. A high tension (HT) supply of, say, 6 kV must have a large internal resistance. Why?

Answer - 18 : - 1. The current will be constant because it is given to be steady.

2.    Ohm’s law is not a fundamental law in nature. It is notuniversally followed. Semiconductor diodes, transistors, their mistors, vacuumtubes etc. do not follow Ohm’s law.

 3.  If emf of supply battery isE and internal resistance r, then current through an external resistance R isgiven by                        so, internal resistance r should be least to supply highcurrents.
4.    In high tension supply, theinternal resistance is made large. Because, if accidently the short circuitingtake place, the excessive current produced should not cross the safety limits.

Question - 19 : -

Choose the correctalternative:

  1. Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
  2. Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
  3. The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
  4. The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022 or 103).

Answer - 19 : -

  1. Alloys of metals usually have greater resistivity than that of their constituent metals.
  2. Alloys usually have much lower temperature coefficients of resistance than pure metals.
  3. The resistivity of the alloy manganin is nearly independent of increasing temperature.
  4. The resistivity of a typical insulator (e.g. amber) is greater than that of a metal by factor of the order of 1022.

Question - 20 : -

(a) Given nresistors each of resistance R, how will you combine them to get the

  • maximum
  • minimum effective resistance? What is the ratio of the maximum to minimum resistance?

(b) Given theresistances of 1 Ω, 2Ω, 3Ω, how will be combine them to get an equivalentresistance of

  • (11/3) Ω
  • (11/5) Ω,
  • 6 Ω,
  • (6/11) Ω?

(c) Determinethe equivalent resistance of network shown in figure.

Answer - 20 : - (a)
For maximum effective resistance, all the resistors should be joined in series. Rmmax = R + R + R+……..n or Rmax = nR
For minimum effective resistance, all the resistors should be joined in parallel.

(b) All possible combinations with resistances 1 Ω, 2 Ω and 3 Ω are



So, the combinations for the desired results can be selected.

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