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 \(100 \mathrm{~V}\)?
- A \(5975 \Omega\)
- B \(20050 \Omega\)
- C \(19950 \Omega\)
- D \(19500 \Omega\)
Answer & Solution
Correct Answer
(C) \(19950 \Omega\)
Step-by-step Solution
Detailed explanation
R= \(\frac{V}{I_g}-R_g=\frac{100}{5 \times 10^{-3}}-50\) \(=20000-50\) \(=19950 \Omega\)
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