Electricity

MCQs Part 2

21. Two conductors made of same material have lengths L and 2 L, but have equal resistances. The two are connected in series in a circuit in which current is flowing. Which of the following is/are correct?

  1. The electron drift velocity is larger in the conductor of length L.
  2. The potential difference across the two conductors is the same.
  3. The electric field in the second conductor is twice that in the first.
  4. The electric field in the first second conductor is twice that in the second.

Answer: Since both the conductors are of equal resistance and in series they carry equal current, hence potential difference across the two conductors is the same. The area of corss-section of the two conductors of length L and 2 L will be A and 2A, for their same resistances. As

E = V/L

22. Two cells of unequal emfs E1 and E2 and internal resistance r1 and r2 are joined as shown in fig. Vp and Vq are the potential at P and Q respectively.

  1. The potential difference across both the cells will be equal
  2. One cell will continuously supply energy to the other
  3. The potential difference across one cell will be greater than its emf.

Answer: Let E1 > E2 . Current in the circuit,

across each cell.

Here, therefore current flows in E2 from the positive plate to the negative plate inside the cell. It means, there is a charging of cell E2 by cell E1.

23. Three ammeters A1, A2 and A3 of resistances R1, R2 and R3 respectively are joined as shown. When some potential difference is applied across the terminals P and Q, their readings are I1 ,I2 and I3 respectively:

Answer: As A1 and A2 are in series, so current in each of them is same i.e. I1 = I2.

Pot. diff. across P and Q

= (I1 R1 + I2 R2) = I3 R3.

= I2 R1 + I2 R2 = I3 R3

24. Two cells of equal emf and having different internal resistance r1 and r2 (r2 > r1)are connected in series. If the resistance of connecting wire is R; which of the following statement is/are correct?

  1. If R = 0, negative terminal of second cell will be at higher potential than its positive terminal.
  2. At a particular value of R, potential difference across second cell can be equal to zero.
  3. Negative terminal of first cell can never be at higher potential than its positive terminal.
  4. None of the above.

Answer: If E is the emf of each cell, then current in the circuit,

Pot. diff. across second cell,

V2 = E—Ir2

If V2 = 0,

then 0 = E — Ir2 or E = Ir2

Putting value in (i), we have

On solving, R = r2—r1

When R = 0 and cells are series, since r2 > r1 so the negative terminal of second cell will be at higher potential than its positive terminal and negative terminal of first cell can never be at higher potential than its positive terminal.

25. A cell drives a current through a circuit : The emf of the cell is equal to the work done in moving unit charge

  1. from the positive plate back to the positive plate
  2. from the positive plate to the negative plate of the cell
  3. from the negative plate, back to the negative plate
  4. from any point in the circuit back to the same point.

Answer: The emf of a cell is equal to the work done in moving unit charge once around a closed circuit starting from any point and back to the same point.

26. The fig. shows a potenetiometer arrangement. B2 is the driving cell. B1 is the cell whose emf is to be determined. AB is the potentiometer wire and G is a galvanometer. J is a sliding contact, which can touch any point on AB. Which of the following are essential conditions for obtaining balance?

  1. The emf of B2 must be greater than the emf of B1
  2. Either the positive terminals of both B1 and B2 or the negative terminals of both B1 and B2 must be joined at A
  3. The positive terminal of B2 must be joined to A
  4. The resistance of G must be less than the resistance of wire AB.

Answer: The balance point can be obtained on the potentiometer wire (i) if the emf of driving cell B2 is greater than the emf of driven cell B1. (ii) either the positive terminals of both the cells of negative terminals of both the cells B1 and B2 must be joined at A.

27. In the potentiometer arrangement shown, the driving cell B1, whose emf is to be measured, has emf E and internal resistance r. The cell B1, whose emf is to be measured, has emf E/2 and internal resistance 2 r. The potentiometer wire is 100 cm long. If balance is obtained, the length AJ = l. Then—

  1. l = 50 cm
  2. l > 50 cm
  3. balance will be obtained only if resistance of wire AB is greater than r
  4. balance cannot be obtained

Answer: If balance is obtained at length l, and R is the resistance of potentiometer wire of length 100 cm, then

So balancing length is greater than 50 cm. If l is greater than 50 cm, then R > r.

28. In the circuit shown fig. in the current

  1. through the 3W resistance is 1 A
  2. through the 3W resistance is 0.5 A
  3. through the 4W resistance is 0.5 A
  4. through the 4W resistance is 0.25 A

Answer: Let the distribution of current in various arms be as shown in fig.

In closed circuit ABHJA

In closed circuit CDFGC

In a closed circuit BCGHB

From (1) and (3),

From (2),

29. Three voltmeters, all having different resistances, are joined as shown. When some potential difference is applied across P and Q, their readings are V1, V2 and V3 respectively. Then—

Answer: Pot. diff. across P and Q = Pot. diff. across V1 and V2 = Pot. diff. across V3

i. e. V1 + V2 =V3

It means V1 V2

30. A voltmeter and an ammeter are connected in series to an ideal cell of emf E. The voltmeter reading is V and the ammeter reading is I.

  1. The voltemeter resistance is V/I
  2. The potential difference across the ammeter (E—V)
  3. V < E
  4. Voltmeter resistance plus ammeter resistance = E/I

Answer: Consider voltmeter and ammeter as resistors connected in series to an ideal cell of emf E. Here pot. diff. across voltmeter is V and current through circuit is I.

(i) The voltmeter resistence = V/I

(ii) Pot. diff. across ammeter = E—V

(iii) E > V

(iv) Resistance of circuit = Voltmeter resistance plus ammeter resistence = E/I.

31. Three resistance of values 2W, 3W and 6 W can be connected to give an effective resistance of 4W.

Answer: It is so when a parallel combination of 3W and 6W resistances are connected in series to 2W resistance.

32. The resultant resistance value of n parallel resistances each of r ohm is x. When these n resistances are connected in series, the resultant value is r n x. True or False Statements

Answer: In parallel combination, total resistance, x = r/n or r = xn. In series combination, total resistance = nr = n (xn) = n2 x.

33. S1,S2,S3 are the conductances of three conductors. When they are joined in series, their equivalent conductance will be (S1+ S2+ S3)?

Answer: Resistance of these conductors will be R1 = 1/S1, R2 = 1/S2 and R3 = 1/S3. In series combination, total resitance

Total conductance,

34. A cell of emf E and internal resistance r if connected in series with an external resistance n r. Then the ratio of terminal potential difference to

Answer: Current in circuit,

Terminal potential difference,

35. Wheatstone bridge is most sensitive, when resistance of all the four arms of the bridge is same.

Answer: Wheatstone bridge is most sensitive, when P = Q = R = S.

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