Question

A current of 1 ampere flows in series circuit contains an electric lamp and a conductor of 5 ohm when connected to a 10V battery. Calculate the resistance of the 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.

Answer 1

I and v are given
R=v/I
R=10/1
R=10
Now total r = 10ohm
Therefore, r of conductor + r of lamp = 10 ohm
5 + r of lamp = 10ohm
R of lamp = 10-5 =5

Answer 2

Answer:

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Total resistance of circuit can be calculated as follows:

$R=\frac{V}{I}$

$=\frac{10V}{1A}$

=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