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When a galvanometer is shunted with a 4 \Omega resistance, the deflection is reduced to 1/5. If the galvanometer is further shunted with a 2\Omega wire, the new deflection will be (assuming the main current remains the same)
 

Option: 1

\frac{5}{13}  of the deflection when shunted with 4\Omega only


Option: 2

\frac{8}{13} of the deflection when shunted with 4\Omega only


Option: 3

\frac{3}{4} of the deflection when shunted with 4\Omega only


Option: 4

\frac{3}{13} of the deflection when shunted with 4 \Omega only


Answers (1)

best_answer

When only 4\Omega resistance is shunted

\mathrm{\left(i_{\mathrm{g}}\right)=i / 5 G \times i / 5=4 \times(4 / 5) \Rightarrow G=16 \Omega}

Posted by

Ajit Kumar Dubey

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