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

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Solution

Let the resistance of the lamp be r. When the conductor of resistance 4 Î© is in series with the lamp, the resistance of the circuit is, R = r+4
Now, the current drawn is 0.25 A from a battery of 6 V.
Thus, 6 = 0.25(r+4)
=> r = 20 Î©

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Now if a resistance of 10 $Ω$ is connected in parallel with this series combination, what change (if any) in current flowing through 5 $Ω$ conductors and the potential difference across the lamp will take place? Give reason.