PUMPA - SMART LEARNING

எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

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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.
 
Let \(R_{1}\) be the resistance of electric lamp. Then, the total resistance is
 
R=ii+i
 
By Ohm's law , we get the value \(R_1\) as  Ω.
 
The resistance of the lamp is connected in series, then \(R\) \(=\) Ω. Then,
 
1RP=iiR=iΩ
 
The value of current is given by the formula
 
I=ii
 
On substituting the known values, we get  \(A\).