A current of 500 mA flows in a series circuit containing an electric
lamp and a conductor of 10 12 when connected to 6 V battery. find
the resistance of the electric lamp.
Answers
Explanation:
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=R1+R2 where R2 is the resistance of the electric lamp.
That is, R=5+R2
Substituting this in the ohm's law R=IV, we get 110=5+R2
So, R2 =5 ohms.
The resistance of the electric lamp is 5 ohms.
The total resistance across the circuit = R1+R2=5+5=10ohms.
When a resistance of 10 ohms is connected in parallel with this series combination, the total resistance through the circuit is R1=101+101=5ohms.
The current across the circuit is I=RV=510=2A
The potential difference across the metallic conductor of 5 ohms is V=IR=2×5=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.