A uniform resistance wire XY of length 100 cm is connected in series with a cell L. Another cell M is connected in series with resistors of resistances 5.00 Ω, 10.0 Ω and 15.0 Ω.

Question 9

A uniform resistance wire XY of length 100 cm is connected in series with a cell L. Another cell M is connected in series with resistors of resistances 5.00 Ω, 10.0 Ω and 15.0 Ω.

The potential difference (p.d.) between P and Q is balanced against 12.5 cm of the resistance wire, so that the ammeter reads zero.

The p.d. across the other resistors is then balanced against other lengths of the resistance wire.

Which balanced lengths of resistance wire correspond to the connection points given in the table?

Reference: Past Exam Paper – June 2018 Paper 11 Q36

Solution:

Answer: B.

For the wire:

Resistance of the resistive wire is given by : R = ρL/A

Since the wire is uniform, R is proportional to L.

From Ohm’s law, V = IR        that is, V is proportional to R.

Combining the 2 relations, we can say that the p.d. V is proportional to L.

For the first resistor, V is proportional to 12.5 cm

At the balanced length, the p.d. between P and Q is the same as the p.d. between X and the position of the junction.

For the resistors:

Ohm’s law: V = IR     that is, V is proportional to R

For PQ, V is proportional to 5.00 Ω

Again, for the ammeter to read zero (balance), the p.d. across the resistor should be equal to the p.d. across X and the position of the junction.

5.00 Ω - - > 12.5 cm

1.00 Ω - - > (12.5 / 5) cm

Across Q and R, resistance = 10.0 Ω

10.0 Ω - - - > 12.5×10 / 5 = 25 cm

Across Q and S, resistance = 10.0 + 15.0 = 25.0 Ω

25.0 Ω - - - > 12.5×25 / 5 = 62.5 cm

Across P and R, resistance = 5.00 + 10.0 = 15.0 Ω

15.0 Ω - - - > 12.5×15 / 5 = 37.5 cm