(a) A current of `1 A` flows in a series curcuit containing an electric lamp and a conductor of `5 Omega` when connected to a `10 V` battery. Calculate the resistance of the electric lamp.
(b) Now if a resistance of `10 Omega` is connected in parallel with this series combination, what change (if any) in current flowing through `5 Omega` conductor and potential difference across the lamp will take place ? Give reason.

Explanation

(i) Let R be the resistance of the electric lamp. In series total resistance = 5+R`I=(V )/(R )`
`1=(10)/(5+R)`
`R=5 omega`
(ii)Total Resistance `=(1)/(R_("eff"))=1/10+1/5+5`
`=1/10+1/10`
`I=(V )/(R )=10/5=2` Amp
Current in each branch =`2/2` = 1 Amp (since, both branches have same resistances, current divides equally) V across lamp + conductor = 10 V V across lamp = `1 xx R = 1 xx 5 = 5` volt