In the circuit shown, sliding contact is moving with uniform velocity towards right. Its value at some instance is 12 Ω. The current in the circuit at this instant of time will be

Since resistance of the circuit is increasing and hence current in the circuit is decreasing, so didt is negative. Current in the circuit is given by i=6-Ldidt12. Since didt is negative so the value of numerator will be more than 6 V and hence current in the circuit at that instant will be more than 0.5 A.