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 -

Pick an option to check your answer.
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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