Question

A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5Ω when connected to a 10V 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 5Ω conductor and potential difference across the lamp will take place? Give reason

Answer 1

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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

Answer 2

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

I = V/ R

1 = 10/ 5 + R

R = 5 ohm

Part 2:

Total Resistance = 10 * 10 / 10 + 10 = 5 ohm

I = V/R = 10/ 5 = 2amp

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

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