JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A \(50\,\Omega \) resistance is connected to a battery of \(5\,V\). A galvanometer of resistance \(100\, \Omega \) is to be used as an ammeter to measure current through the resistance, for this a resistance \(r_s\) is connected to the galvanometer. Which of the following connections should be employed if the measured current is within \(1\% \) of the current without the ammeter in the circuit ?
- A \(r_s = 0.5\, \Omega \) in series with the galvanometer
- B \(r_s = 1\, \Omega \) in series with galvanometer
- C \(r_s = 1\, \Omega \) in parallel with galvanometer
- D \(r_s = 0.5\, \Omega \) in parallel with the galvanometer
Answer & Solution
Correct Answer
(D) \(r_s = 0.5\, \Omega \) in parallel with the galvanometer
Step-by-step Solution
Detailed explanation
As we know, \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{5}{50}=0.1\) \(I' = 0.099\) When Galvanometer is connected \(R_{e q}=50+\frac{100 S}{100+S}=\frac{V}{I}\) \(\Rightarrow \frac{100 \mathrm{S}}{100+\mathrm{S}}=\frac{5}{0.099}-50\)…
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