Q153 / 180 · KCET 2016 · Physics
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
current of \( 5 \mathrm{~mA} \), This galvanometer is converted into voltmeter of range \( 0-20 \mathrm{~V} \) by connecting
- Correct answerA\( 3950 \Omega \) 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
Solution
We know, \( R=\frac{V}{I_{g}}-G \)
Given,
\[ \begin{array}{l} V=20 \mathrm{~V} ; I_{g}=5 \mathrm{~mA}=5 \times 10^{-3} \mathrm{~A} ; G=50 \Omega \\ \Rightarrow R=\frac{20}{5 \times 10^{-3}}-50=\frac{20}{0.005}-50=3950 \Omega \end{array} \]
Therefore, the given galvanometer is converted into voltmeter of range \( 0-20 \mathrm{~V} \) by connecting \( 3950 \Omega \) in series with
galvanometer.
Given,
\[ \begin{array}{l} V=20 \mathrm{~V} ; I_{g}=5 \mathrm{~mA}=5 \times 10^{-3} \mathrm{~A} ; G=50 \Omega \\ \Rightarrow R=\frac{20}{5 \times 10^{-3}}-50=\frac{20}{0.005}-50=3950 \Omega \end{array} \]
Therefore, the given galvanometer is converted into voltmeter of range \( 0-20 \mathrm{~V} \) by connecting \( 3950 \Omega \) in series with
galvanometer.
Hence, the correct answer is option: 'A'.