Get Answers to all your Questions

header-bg qa

When a galvanometer is shunted with a 4\Omega resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2\Omega wire, the further reduction (find the ratio of decrease in current to the previous current) in the deflection will be (the main current remains the same).

Option: 1

(8/13) of the deflection when shunted with 4\Omega only


Option: 2

(5/13) of the deflection when shunted with 4\Omega only


Option: 3

(3/4) of the deflection when shunted with 4\Omega only


Option: 4

(3/13) of the deflection when shunted with 4\Omega only


Answers (1)

best_answer

 

Conversion of galvanometer into ammeter -

Connect a low resistance in parallel to the galvanometer

 

Case I

R_{g}\times \frac{I}{5}=\left ( I-\frac{I}{5} \right )\times 4\Rightarrow R_{g}=16\Omega

Case II

16I_{1}=\frac{4 \times 2}{6}(I-I_{1})\Rightarrow I_{1}=I/13

so decrease in current to the previous current

=\frac{I/5-I/13}{I/5}=\frac{8}{13}

Posted by

Deependra Verma

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE