Question

A conductor of resistance 12.8ohm is cut into a number of equal parts. All these parts are connected in parallel. The resistance of the combination so obtained is 0.2 ohm. The number of parts are

Answer 1

A conductor of resistance 12.8ohm is cut into a number of equal parts. All these parts are connected in parallel. The resistance of the combination so obtained is 0.2 ohm. The number of parts are

Let say Conductor is cut into n equal parts

Resistance of One Part = 12.8/n

these n resistances are connected in Parallel so

net resistance of combination

=   1/( 1/(12.8/n)   + 1/(12.8/n)  + ......n times)

= 1/ (n + n +   ..... n times)/12.8)

= 12.8/n²

12.8/n² = 0.2

=> n² = 12.8/0.2

=> n² = 64

=> n = 8

Wire was cut in 8 equal parts

each having resistance of 12.8/8 = 1.6 Ω