COMEDK · Physics · 19. Current Electricity
In an ammeter \(10 \%\) of main current is passing through the galvanometer. If the resistance of the galvanometer is \(G\), then the shunt resistance (in \(\Omega\) ) is
- A \(9 G\)
- B \(\frac{G}{9}\)
- C \(90 G\)
- D \(\frac{G}{90}\)
Answer & Solution
Correct Answer
(B) \(\frac{G}{9}\)
Step-by-step Solution
Detailed explanation
If \(I\) be the main current, then current passing through galvanometer, \(I_{g}=10 \% \text { of } I=\frac{I}{10}\) Resistance of galvanometer, \(R_{g}=G\) Shunt resistance, \(R_{S}=\frac{I_{g} R_{g}}{I-I_{g}}=\frac{\frac{I}{10} \times G}{I-\frac{I}{10}}=\frac{G}{9}\)
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