Q26 of 35 Page 93

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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