As electric lamp and a conductor of resistance `4 Omega` are connected in series to a 6 V battery. The current drawn by the lamp is 0.25 A. Find the resistance of the electric lamp.

Explanation

Lamp `:`
Resistance , `R_(1) = ?`
Potential differences, `V_(1) = ?`
Current drawn, `I_(1)= 0.25 A`
Other conductor `:`
Resistance , `R_(2) = 4 Omega`
Potential differences , `V_(2) = ?`
Current drawn , `I_(2) =?`
Resultant resistance, `R =R_(1)+R_(2)`
`:.` Potential differene, `V = 6V`
Current in the circuit , `I = 0.25A`
`:' I= I_(1)= I_(2)`
`:. V= IR`
`implies R = (V)/(I) = (6)/(0.25) = (6 xx cancel(100)^(4))/(cancel(25))=24 Omega`
Now, `R_(1)+R_(2)=R` `implies R_(1)+4=24`
`:. R_(1)= 24-4= 20 Omega`