JEE Mains · Physics · STD 12 - 3. current electricity
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 -
- A \(\frac{40}{11}\, {V}\)
- B \(10\, {V}\)
- C \(\frac{40}{9} \,{V}\)
- D \(5 \,{V}\)
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…
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