A current of 1 ampere flows in a series circuit having 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
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=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=
I
V
, we get
1
10
=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
R
1
=
10
1
+
10
1
=5ohms.
The current across the circuit is I=
R
V
=
5
10
=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.