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JEE Mains · Physics · STD 12 - 3. current electricity

Four resistances of \(15\; \Omega, 12\; \Omega, 4 \;\Omega\) and \(10\; \Omega\) respectively in cyclic order to form Wheats tone's network. The resistance that is to be connected in parallel with the resistance of \(10\; \Omega\) to balance the network is .................. \(\Omega\)

  1. A \(13\)
  2. B \(10\)
  3. C \(7\)
  4. D \(17\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(10\)

Step-by-step Solution

Detailed explanation

Let the resistance to be connected is \(\mathrm{R}\). For balanced wheatstone bridge. \(15 \times 4=12 \times \frac{10 \mathrm{R}}{10+\mathrm{R}}\) \(\Rightarrow \mathrm{R}=10 \Omega\)
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