KCET · Physics · Current Electricity
A galvanometer coil has a resistance of \( 50 \Omega \) and the meter shows full scale deflection for a current of \( 5 \mathrm{~mA} \), This galvanometer is converted into voltmeter of range \( 0-20 \mathrm{~V} \) by connecting
- A \( 3950 \Omega \) in series with galvanometer
- B \( 4050 \Omega \) in series with galvanometer
- C \( 3950 \Omega \) in parallel with galvanometer
- D \( 4050 \Omega \) in parallel with galvanometer
Answer & Solution
Correct Answer
(A) \( 3950 \Omega \) in series with galvanometer
Step-by-step Solution
Detailed explanation
We know, \( R=\frac{V}{I_{g}}-G \)
Given,
\(V=20 \mathrm{~V} ; I_{g}=5 \mathrm{~mA}=5 \times 10^{-3} \mathrm{~A} ; G=50 \Omega \)
\(\Rightarrow R=\frac{20}{5 \times 10^{-3}}-50=\frac{20}{0.005}-50=3950 \Omega\)
Therefore, the given galvanometer is converted into voltmeter of range \( 0-20 \mathrm{~V} \) by connecting \( 3950 \Omega \) in series with
galvanometer.
Given,
\(V=20 \mathrm{~V} ; I_{g}=5 \mathrm{~mA}=5 \times 10^{-3} \mathrm{~A} ; G=50 \Omega \)
\(\Rightarrow R=\frac{20}{5 \times 10^{-3}}-50=\frac{20}{0.005}-50=3950 \Omega\)
Therefore, the given galvanometer is converted into voltmeter of range \( 0-20 \mathrm{~V} \) by connecting \( 3950 \Omega \) in series with
galvanometer.
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