A heater of resistance `140 Omega` capable of carrying a current of `1.2 A` is put in a dc mains of 210 V. Find out the minimum value of an additional resistance to be added to run the heater.

Explanation

Let the value of the minimum additional resistance to be added be x `Omega`.
We know, V=IR
or, `210=1.2(140+x)[Here R=140+x]`
`or,140+x=210/1.2or,x=175-140or,x=35Omega`
So, a minimum additional resistance of `35 Omega` is to be added to run the heater.