A current of 1 ampere flows in series circuit contains an electric lamp and a conductor of 5 ohm when connected to a 10V battery. Calculate the resistance of the electric lamp.

Now if a resistance of 10 ohm is connected in parallel with this series combination, what change in current flowing through 5 ohm conductor and potential difference across the lamp will take place? Give reason.

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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
2 4 2
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