COMEDK · Physics · 19. Current Electricity
The maximum current that can be measured by a galvanometer of resistance \(40 ~\Omega\), is \(10 \mathrm{~mA}\). It is converted into a voltmeter that can read upto \(50 \mathrm{~V}\). The resistance to be connected in series with the galvanometer (in \(\Omega\) ) is
- A \(2010\)
- B \(4050\)
- C \(5040\)
- D \(4960\)
Answer & Solution
Correct Answer
(D) \(4960\)
Step-by-step Solution
Detailed explanation
Given, \(G=40 \Omega, I_{g}=10 \mathrm{~mA}=10 \times 10^{-3} \mathrm{~A}\) \[ V=50 \mathrm{~V} \] Let \(R\) be the resistance connected in series with the galvanometer to convert into voltmeter, then…
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