Question

A current of 1 ampere flows in a series circuit containing 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

Answer 1

I=1A
V=10V
R=10/1=10Ω=5+LAMP
LAMP=r=5Ω
there would be no change in the potantial difference as in a parallel combination it remains same
about current,
the total current flowing through the circuit=10/5=2A
this current is divided between two 10Ω in parallel connection
so,each part of the parallel connection gets 1A each and this is the current that flows through the the 5Ω resistor

Answer 2

Answer:

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