(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

(a) Total resistance in the circuit, `R = (V)/(I) = (10 V)/(1 A) = 10 Omega`
Since conductor of `5 Omega` and lamp are in series, `R = 5 Omega + R_("lamp"),R_("lamp") = R -5 Omega = 10 Omega - 5 Omega = 5 Omega`
pd across lamp `= I(R_("lamp") = (1 A)(5 Omega) = 5 V`
(ii) When `10 Omega` resistance is connected in parallel with `R(=10 Omega)`,
total resistance `(R')` in the circuit is given by
`(1)/(R') = (1)/(10)+(1)/(R )= (1)/(10)+(1)/(10) = (1)/(5)` or `R' = 5 Omega`.
Current through the circuit, `I' = (V)/(R') = (10 V)/(10 Omega) = 2 A`
Since `10 Omega` and `R(=10 Omega)` are in parallel, current through `R i.e., (I')/(2) = (2 A)/(2) = 1 A`
Thus, current through lamp and conductor of `5 Omega` in series is `1 A` i.e., there is no change in current through conductor of `5 Omega`.
Further, pd across lamp `= (I')/(2) (R_("lamp")) = (1 A) (5 Omega) = 5 V`, i.e., there is no change in pd across the lamp.