Question

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. The resistance of the electric lamp will be :

• A. 1 $\Omega$
• B. 6 $\Omega$
• C. 5 $\Omega$
• D. 10 $\Omega$

Answer 1

The correct option is C 5 $\Omega$

In the question, it is given that the resistance of the conductor is 5 ohms and the current flowing through the circuit is 1A and the potential difference is 10V. The total resistance R is given as

$R=R_1+R_2$ where $R_2$ is the resistance of the electric lamp.
That is, $R=5+R_2$

Substituting this in the ohm's law $R=\frac{V}{I}$, we get $\frac{10}{1}=5+R_2$
So, $R_2$ =5 ohms.

The resistance of the electric lamp is 5 ohms.
The total resistance across the circuit = $R_1+R_2=5+5=10ohms$.
When a resistance of 10 ohms is connected in parallel with this series combination, the total resistance through the circuit is $\frac{1}{R}=\frac{1}{10}+\frac{1}{10}=5ohms$.
The current across the circuit is $I=\frac{V}{R}=\frac{10}{5}=2A$
The potential difference across the metallic conductor of 5 ohms is $V=IR=2\times 5=10V.$
Hence, there will be no change in current flowing through 5 ohms conductor ,also there will be no change in potential difference across the lamp either.