A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 ohm when connected to a 10v battery. calculate the resistance of electric lamp. now if a resistance of 10 ohm is connected in parallel with this series combination, what change in current flowing through 5 ohm conductor and potential difference across the lamp will take place? give reason
Answers
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 }R=R
1
+R
2
where { R }_{ 2 }R
2
is the resistance of the electric lamp.
That is, R=5+{ R }_{ 2 }R=5+R
2
Substituting this in the ohm's law R=\dfrac { V }{ I }R=
I
V
, we get \dfrac { 10 }{ 1 } =5+{ R }_{ 2 }
1
10
=5+R
2
So, { R }_{ 2 }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=10\quad ohmsR
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 \dfrac { 1 }{ R } ={ \dfrac { 1 }{ 10 } }+\dfrac { 1 }{ 10 } =5\quad ohms
R
1
=
10
1
+
10
1
=5ohms.
The current across the circuit is I=\dfrac { V }{ R } =\dfrac { 10 }{ 5 } =2AI=
R
V
=
5
10
=2A
The potential difference across the metallic conductor of 5 ohms is V=IR=2\times 5=10V.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
Answer:
this is from NCERT example I have solved it for you below
Total resistance of circuit can be calculated as follows:
=10ohm
Since lamp and conductor are in series so resistance of lamp
10ohm-5ohm=5ohm
The new resistance in parallel to earlier combination has same value, i.e. 100 as the resistance of series combination. This means that the amount of current would be equally divided into two branches. Hence, 0.5A current will flow through 30 conductor
Now, resistance remains the same but current has become half. Using Ohm formula,
potential difference across the lamp can be calculated as follows:
V=IR=0.5A×1ohm=2.5V