Question

You are supplied with a battery that has an emf of 10V and a piece of rubber with a resistivity of 8x1013 ohm m. The rubber has dimensions of 2 cm x 3 cm x 9 cm.

A. If you wanted to use the piece of rubber as a resistor with the lowest possible resistance which faces would you connect the battery to? The 2x3 cm2 , 2x9 cm2 , or 3x9 cm2 sides? Why?

B. Calculate the resistance of the piece of rubber if the battery is connected to the 2x3 cm2 faces. This will be the ‘resistor’ for the following steps.

C. Sketch the arrangement of the battery connected to the 2x3 cm2 ends of the rubber, the resistor.

D. When the resistor is connected to the battery at the 2x3cm2 ends, what is the current through the resistor?

E. What is the potential change across the resistor?

F. What is the power dissipated in the resistor?

G. If an identical battery is connect to the first one, what is the emf and the current in the circuit?

Answer 1

A) we know, R = rho*L/A

L --> length
A --> cross sectional area

so, for lowest resistance we take A = 3*9 cm^2 , L = 2 cm

B)

R = rho*L/A

= 8*10^13*9*10^-2/(2*3*10^-4)

= 1.2*10^16 ohms

C)

D) I = V/R

= 10/(1.2*10^16)

= 8.33*10^-16 A

E) Potential change across the resistor = 10 V

F) Power dissipated, P = I^2*R

= (8.33*10^-16)^2*1.2*10^16

= 8.33*10^15 W

G) R = rho*L/A

= 8*10^13*2*10^-3/(3*9*10^-4)

= 5.92*10^13 ohms

I = V/R

= 10/(5.92*10^13)

= 1.67*10^-13 A