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) Let R1 be the resistance of the electric lamp.

R2 be the resistance of the conductor.

In series total resistance = R1 + R2 = 5 + R1

Current(I) = 1 A

Voltage (V) = 10V

Using ohms law

I = V/R

1 = 10/(5+R)

5 + R = 10

R1 = 10 – 5

R = 5Ω

(2) Now, a resistance of 10Ωis connected in parallel with the series combination. Therefore, the total resistance of the circuit is given by

1/Rp = 1/(R1 + 5) + 1/10

1/Rp = 1/10 + 1/10

Rp = 5Ω

Hence current flowing in the cuircuit

I = V/R

I = 10/5

I = 2A

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