28.A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5ohm when connected to a 10V
battery. Calculate the resistance of the electric lamp Now a resistance of 10 ohm is connected in parallel tanis series
combination what change (if any) in current flowing through 5ohm conductor and potential atence across the lamp will tale
place? Give reason
Answers
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=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...
Mark me as BRAINLIST