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Magazine Chapter 21: Q45PE (page 777)
Find the resistance that must be placed in series with a \({\rm{25}}{\rm{.0}}\;{\rm{\Omega }}\) galvanometer having a \({\rm{50}}{\rm{.0}}\;{\rm{\mu A}}\) sensitivity (the same as the one discussed in the text) to allow it to be used as a voltmeter with a \({\rm{3000}}\;{\rm{V}}\) full-scale reading. Include a circuit diagram with your solution.
Short Answer
The resistance in series with the galvanometer that has inner resistance \({\rm{25}}{\rm{.0}}\;{\rm{\Omega }}\) on the voltmeter鈥檚 \({\rm{3000}}\;{\rm{V}}\) scale and sensitivity of \({\rm{50}}{\rm{.0}}\;{\rm{\mu A}}\), is \(R = 60 {\rm{M\Omega }}\).
Step by step solution
Concept Introduction
When a high resistance is added in series with a galvanometer, the setup can be used as a voltmeter to measure potential difference. The sensitivity of the galvanometer is measured by the deflection it produces per unit change in current.
Information Provided
- Inner resistance in galvanometer:\({\rm{25}}{\rm{.0}}\;{\rm{\Omega }}\)
- Sensitivity of galvanometer:\({\rm{50}}{\rm{.0}}\;{\rm{\mu A}}\)
- Scale measure of voltmeter: \({\rm{3000}}\;{\rm{V}}\)
Calculation for Current
Calculate the additional resistance \(R\) necessary for the galvanometer to have a \(V = 3000\;{\rm{V}}\) full-scale reading. The total resistance in this case is \(r + R\). Calculate \(R\) as 鈥�
\(r + R = \frac{V}{I}\)
\(\begin{array}{c}R = \frac{V}{I} - r\\ = \frac{{3000\;{\rm{V}}}}{{50 {\rm{mA}}}} - 25 \Omega \\ = 60 {\rm{M\Omega }}\end{array}\)
The setup is depicted in the figure below.
Therefore, the value for resistance is obtained as \(R = 60 {\rm{M}}\Omega \).
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