Question

A current of 1 A 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. Draw Circuit diagram

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Answer 1

Part 1:

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

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

1 =  $\frac{10}{5+R}$

R = 5 ohm 

Part 2:

Total Resistance =  $\frac{10 \times 10}{10+10} = 5\ ohm$

I =  $\frac{V}{R}=\frac{10}{5} = 2 Amp$

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