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

Explanation

(i) Let the resistance of the lamp = `R_(1)`
Resistance of conductor = `R_(2) = 5Omega`
Total resistance in series, `R_(S) = R_(1) + R_(2)= R_(1)+5`

Current I = 1 A, Voltage, V = 10 V
Using Ohm.s Law, `V= IR`
`10= 1(R_(1)+5)`
`implies R_(1)= 5Omega`
(ii) Now, a resistance of `10Omega` is connected in parallel with the series combination. Therefore, the total resistance of the circuit is given by
`(1)/(R_(P))= (1)/(R_(1)+5)+(1)/(10)`
`(1)/(R_(P))= (1)/(10)+ (1)/(10)`
`R_(P)= 5Omega`

Hence, current flowing in the circuit,
`I= (V)/(R)= (10)/(5)= 2A`
Thus, 1 A current will flow through `10Omega` resistor and 1 A will flow through the lamp and conductor of `5Omega` resistance. Hence, there will be no change in current flowing through `5Omega` conductor. Also, there will be no change in potential difference across the lamp.