# A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. 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 conductor and potential difference across the lamp will take place? Give reason.

When 5Ω and resistance of lamp say R is connected in series:

Effective resistance R’ = 5+R

Applying ohm’s law:

V= IR

R = V/I = 10/1 = 10 Ω

Therefore, resistance of lamp, R= 10-5 Ω = 5 Ω

When a 10 Ω resistance is connected in parallel with the series connection

Effective resistance 1/R’’= 1/10 +1/10

R’’ = 5 Ω

Current through the circuit I = V/R = 10/5 = 2A

So, current flowing through each circuit will be of 1 V

Potential difference across the lamp will be same because voltage is same across the circuit.

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