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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2020
  • Medium
  • Numerical

Four resistances \(40 \ \Omega, 60\  \Omega, 90\  \Omega\) and \(110\  \Omega\) make the arms of a quadrilateral \(A,B,C,D\). Across \(AC\) is a battery of emf \(40\, V\) and internal resistance negligible. The potential difference across \(BD\) is \(V\) is...... 

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Answer & Solution

Correct Answer

(C) \(2\)

Step-by-step Solution

Detailed explanation

\(i _{1}=\frac{40}{40+60}=0.4\)

\(i _{2}=\frac{40}{90+110}=\frac{1}{5}\)

\(v _{ B }+ i _{1}(40)- i _{2}(90)= v _{ D }\)

\(v _{ B }- v _{ D }=\frac{1}{5}(90)-\frac{4}{10} \times 40\)

\(v _{ B }- v _{ D }=18-16=2\)

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