JEE Mains · Physics · STD 12 - 3. current electricity
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......
- A \(4\)
- B \(1\)
- C \(2\)
- D \(5\)
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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