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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2021
  • Hard
  • MCQ

the given potentiometer has its wire of resistance \(10\, \Omega\). When the sliding contact is in the middle of the potentiometer wire, the potential drop across \(2\, \Omega\) resistor is -

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Answer & Solution

Correct Answer

(C) \(\frac{40}{9} \,{V}\)

Step-by-step Solution

Detailed explanation

\(\frac{20- V _{0}}{5}+\frac{0- V _{0}}{5}+\frac{20- V _{0}}{2}=0\)

\(4+10=\frac{2 V _{0}}{5}+\frac{ V _{0}}{2}\)

\(14=\frac{4 V _{0}+5 V _{0}}{10}\)

\(V _{0}=\frac{140}{9} Volt \)

Potential difference across \(2 \Omega\) resistor is \(20- V _{0}\)

That is \(\left(20-\frac{140}{9}\right)\) Volt

Hence answer is \(\left(\frac{40}{9}\right)\) Volt

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