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

A cell, shunted by a \(8 \; \Omega\) resistance, is balanced across a potentiometer wire of length \(3 \;  m\). The balancing length is \(2 \; m\) when the cell is shunted by \(4 \; \Omega\) resistance. The value of internal resistance of the cell will be \(\dots \; \Omega .\)

  1. A \(7\)
  2. B \(8\)
  3. C \(9\)
  4. D \(10\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(8\)

Step-by-step Solution

Detailed explanation

\(\frac{V_{1}}{V_{2}}=\frac{3}{2}=\frac{E-i_{1} r}{E-i_{2} r}\) \(=\frac{E-\frac{E}{8+r} \times r}{E-\frac{E}{4+r} \times r}\) \(\frac{3}{2}=\frac{8(4+r)}{4(8+r)}\) \(24+3 r=16+4 r\) \(r=8 \; \Omega\)
Same subject
Explore more questions on app