JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
When a current of \(5\ mA\) is passed through a galvanometer having a coil of resistance \(15\ \Omega\), it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into to voltmeter of range \(0 - 10\ V\) is
- A \(1.985 \times 10^3\) \(Ω\)
- B \(2.045 \times 10^3 \) \( Ω\)
- C \(2.535 \times 10^3 \) \(Ω\)
- D \(4.005 \times10^3\) \( Ω\)
Answer & Solution
Correct Answer
(A) \(1.985 \times 10^3\) \(Ω\)
Step-by-step Solution
Detailed explanation
Given : Current through the galvanometer, \(i_{g}=5 \times 10^{-3} \,A\) Galvanometer resistance, \(G=15\, \Omega\) Let resistance \(R\) to be put in series with the galvanometer to convert it into a voltmeter. \(V=i_{g}(R+G)\) \(10= 5 \times 10^{-3}(R+15) \)…
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