JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
The current sensitivity of moving coil galvanometer is increased by \(25 \%\). This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be \(...........\%\)
- A \(+25\)
- B \(-50\)
- C \(0\)
- D \(-25\)
Answer & Solution
Correct Answer
(A) \(+25\)
Step-by-step Solution
Detailed explanation
\(I _{ s }=\frac{ NBA }{ C } \& V _{ s }=\frac{ NBA }{ CG }\) \(\Rightarrow V_{ s }=\frac{I_s}{G}\), If \(G\) (galvanometer resistance) is constant, then \(V _{ S } \propto I_S\) so percentage change in \(V_S\) is also \(25 \%\).
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