(i) A current of I A (its capital i, not 1) flows in a series circuit containing an electric lamp and a conductor of 5ohm when connected to a 10 V battery. Calculate resistance of the electric lamp.
(ii)Now If a resistance of 100 is connected in parallel with this series combination, what change (if any) in current flowing through 50 conductor
and potential difference across the lamp will take
place? Give reason?
Answers
ANSWER
I HOPE THIS IS CLEAR
EXPLANATIONS
ANSWER
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.