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

(a) Case I
\begin{aligned} & \mathrm{R}_{\mathrm{g}} \times \frac{\mathrm{I}}{5}=\left(\mathrm{I}-\frac{\mathrm{I}}{5}\right) \times 4 \\ & \Rightarrow \mathrm{R}_{\mathrm{g}}=16 \Omega \end{aligned}
Case II

16 \ \mathrm{I}_1=\frac{4 \times 2}{6}\left(I-\mathrm{I}_1\right) \Rightarrow \mathrm{I}_1=\mathrm{I} / 13
So decrease in current to previous current
=\frac{\mathrm{I} / 5-\mathrm{I} / 13}{\mathrm{I} / 5}=\frac{8}{13}

Posted by

HARSH KANKARIA

View full answer

JEE Main high-scoring chapters and topics

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