Question

The current I through a rod of a certain metallic oxide is given by I = 0.2 V5/2 , where V is the potential difference across it. The rod is connected in series with a resistance to a 6V battery of negligible internal resistance. What value should the series resistance have so that :

(i) the current in the circuit is 0.44 (the value of(2.2)2/5 = 1.37)

(ii) the power dissipated in the rod istwice that dissipated in the resistance.

Answer 1

(i) the potential difference across rod for current 0.44 is

The potential difference across connected resistor is V´´= 0.44 R

(ii)Total power supplied by battery is used by rod and resistor

∴ E I = V I + I2 R

From eqn (i), IR = 2

Remarks: In the case of rod, V– I graph is not straight line. So, ohm’slaw is not applicable in the case of the given rod.