Question

15) A current of 1 ampere flows in a series circuit containing an electric
lamp and a conductor of 5 when connected to a 10 V battery.
Calculate the resistance of the electric lamp.
Now if a resistance of 10
is connected in parallel with this
series combination, what change (if any) in current flowing through
5 n conductor and potential difference across the lamp will take
place? Give reason.

Answer 1

Answer:

Part 1:

Let R be the resistance of the electric lamp. In series total resistance = 5 + R

I =

1 =  

R = 5 ohm 

Part 2:

Total Resistance =  

I =  

Current in each branch  = 2 / 2 = 1 Amp (since both branches have same resistances, current divides equally)

V across Lamp + conductor = 10 V

V acoess Lamp = I × R = 1 * 5 = 5 Volt

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