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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism

A galvanometer has a resistance of \(50 \Omega\) and allows a maximum current of \(5 \mathrm{~mA}\) to pass. What is the necessary series resistance to connect to convert it to a voltmeter that can measure \(100 \mathrm{~V}\)?

  1. A \(5975 \Omega\)
  2. B \(20050 \Omega\)
  3. C \(19950 \Omega\)
  4. D \(19500 \Omega\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(19950 \Omega\)

Step-by-step Solution

Detailed explanation

R= \(\frac{V}{I_g}-R_g=\frac{100}{5 \times 10^{-3}}-50\) \(=20000-50\) \(=19950 \Omega\)