JEE Mains · Physics · STD 12 - 3. current electricity
The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of \(15\, \Omega\) resistance is connected across \(BD\). Calculate the current through the galvanometer when a potential difference of \(10\, V\) is maintained across \(AC.\)
- A \(2.44\, \mu A\)
- B \(2.44\, mA\)
- C \(4.87\, mA\)
- D \(4.87\, \mu A\)
Answer & Solution
Correct Answer
(C) \(4.87\, mA\)
Step-by-step Solution
Detailed explanation
\(\frac{x-10}{100}+\frac{x-y}{15}+\frac{x-0}{10}=0\) \(53 x-20 y=30 \ldots \ldots(1)\) \(\frac{y-10}{60}+\frac{y-x}{15}+\frac{y-0}{5}=0\) \(17 y -4 x =10\) on solving \((1)\,\&\,(2)\) \(x=0.865\) \(y =0.792\) \(\Delta V =0.073 R =15\, \Omega\) \(i =4.87 \,mA\)
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