JEE Mains · Physics · STD 12 - 3. current electricity
A \(16\, \Omega\) wire is bend to form a square loop. A \(9 \,{V}\) supply having internal resistance of \(1 \,\Omega\) is connected across one of its sides. The potential drop across the diagonals of the square loop is \(.......\,\times 10^{-1} \,{V}\)
- A \(45\)
- B \(40\)
- C \(12\)
- D \(33\)
Answer & Solution
Correct Answer
(A) \(45\)
Step-by-step Solution
Detailed explanation
By KVL in outer loop \(9-12 i-4 i=0\) \(16 i=9\) \(8 i =\frac{9}{2}=4.5\) \(=45 \times 10^{-1}\)
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