Question

Question 26
A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 $\Omega$ when connected to a 10 V battery. Calculate the resistance of the electric lamp.
Now if a resistance of 10 $\Omega$ is connected in parallel with this series combination, what change (if any) in current flowing through 5 $\Omega$ conductors and the potential difference across the lamp will take place? Give reason.

Answer 1

According to Ohm's law,
$R=\frac{V}{I}$
$Here,R=\frac{10}{1}=10\Omega$
Since the heater and conductor are in series, resistance of electric lamp = Total resistance- Resistance of the conductor
Hence resistance= $10-5=5\Omega$
Now, when a resistance of 10 Ω is connected in parallel with this series combination, the total resistance is reduced to half of the original value and hence the total current is double the original value.
Therefore, there will be no change in the current flowing through 5$\Omega$ conductors and the potential difference across the lamp will also remain the same.