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

Asked by richaverma914 | 19th May, 2015, 07:27: PM

Expert Answer:

begin mathsize 14px style Let space straight R subscript lamp space be space the space resistance space of space the space electric space lamp Given space that space 5 space straight capital omega space and space straight R subscript lamp space is space connected space in space series colon Let space straight R subscript effective space be space the space total space resistance colon Then space we space have colon straight R subscript effective space equals 5 space straight capital omega space plus straight R subscript lamp space Given space that space the space current space straight I space equals 1 space straight A space and space straight V equals 10 space straight V According space to space ohm apostrophe straight s space law comma straight V equals IR straight R subscript effective equals straight V over straight I equals 10 over 1 equals 10 straight capital omega Therefore space we space get space that colon  straight R subscript lamp space equals straight R subscript effective minus space 5 space straight capital omega straight R subscript lamp space equals 10 space straight capital omega space minus 5 space straight capital omega straight R space subscript lamp space equals 5 space straight capital omega end style
Therefore the resistance of the electric lamp = 5 Ω
 
When a resistance of 10 ohm is connected in parallel with this series combination, the circuit will be as shown below:

begin mathsize 14px style Let space straight R subscript straight p space be space the space total space resistance space of space the space circuit space when 10 straight capital omega space is space connected space in space parallel space with space the space series space combination. 1 over straight R subscript Parallel equals fraction numerator 1 over denominator straight R subscript lamp plus 5 straight capital omega end fraction space plus fraction numerator 1 over denominator 10 space straight capital omega end fraction 1 over straight R subscript Parallel equals fraction numerator 1 over denominator 10 straight capital omega end fraction space plus fraction numerator 1 over denominator 10 space straight capital omega end fraction 1 over straight R subscript Parallel equals fraction numerator 2 over denominator 10 straight capital omega end fraction equals fraction numerator 1 over denominator 5 straight capital omega end fraction rightwards double arrow straight R subscript Parallel space equals 5 space straight capital omega The space current space fowing space through space the space circuit space will space be colon straight I space equals straight V over straight R equals 10 over 5 equals 2 space straight V end style
The current flowing through both the branches will be 1 V each 
i.e. 1 V current flows through 10Ω resistance and 5Ω conductor.
There will be no chnage 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.  
 

Answered by Jyothi Nair | 20th May, 2015, 09:13: AM

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