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  • A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 omega when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 omega is connected in parallel with this

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. 

Answers (1)

The situation can be depicted as:

Current, I = 1A

Resistance of conductor, RC = 5\Omega

Voltage, V = 10V

Resistance of lamp, RL = ?

Total resistance in the circuit, RT = V/ I = (10/1) \Omega = 10 \Omega

As the resistances are in series, RT = RC + RL

RL = 10\Omega – 5\Omega= 5\Omega

Potential difference across the lamp = I RL = (1 × 5) = 5V

Now, 10Ω resistances is connected in parallel with total resistance RT

1/Rnew = (1/10) + (1/10) = 2/10 = 1/5

Rnew = 5 \Omega

New current, Inew = V/Rnew = (10/5) A = 2 A

As both branches have equal resistance, current will divide equally.

Hence, current through 5Ω resistance and lamp = 2/2 = 1 A.

Potential difference across the lamp = 1 × 5 = 5 V.  

There will be no change in the current flowing through 5 ohm conductor and potential difference across the lamp.

As voltage remains the same in both the branches the potential difference across the lamp will also remain the same. 

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infoexpert24

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