# 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.

(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