A current of 1 ampere flows in a series circuit containing 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 `10Omega` 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

Given: In a series circuit containing lamp and resistor,
`I=IA,R_(L)=?R_(C)=5Omega,V=10V`
Using Ohm.s law, `V=1R_(S)`
`V=1[R_(L)+R_(C)]`
`10=1[R_(L)+5]`
`R_(L)=5Omega`.
So, resistance of an electrical lamp is `5Omega`.
According to the given condition, circuit can be redrawn as shown.

So, effective resistance of parallel combination,
`1/R_(P)=1/R_(2)+1/R_(S)`
= `1/R_(2)+1/(R_(L)+R_(1))`
= `1/10+1/(5+5)=1/10+1/10=1/5`
Therefore, `R_(P)=5Omega`
Current drawn from the battery
`I=V/R=10/5=2A`
As in parallel combination, potential difference across them remains same.
So,
`I_(1)R_(AB)=I_(2)R_(CD)`
`rArrI_(1)/I_(2)=R_(CD)/R_(AB)=10/10=1`
`rArrI_(1)=I_(2)`
i.e., Current is divided in both the arms equally. So,
`I_(1)=I_(2)=IA`.
Hence, there will be no change in the current through `5Omega` conductor. Also there will be no change in the potential difference across the lamp as in both cases, current through the lamp remains same i.e. IA.