A `P-N` junction diode can withstand currents up to `10 mA`. Under forward bias, The diode has a potential drop of `0.5 V` across it which is assumed to be independent of current. The maximum voltage of the battery used to forward bias the diode when a resistance of `200 Omega` is connected in series with it is

Explanation

Potential drop is 0.5 and current is 10 mA , so resistance of diode is `(0.5)/(10 xx 10^(-3)) = 50 Omega`
Total resistance of circuit is `200 + 50 = 250 Omega`
So maximum current is `(V_(max))/(250)=10 xx 10^(-3) rArr V_(max) = 2.5` volt.