Get Answers to all your Questions

header-bg qa

To increase the current sensitivity of a moving coil galvanometer by 50% , its resistance is increased by twice the initial value . By what factor voltage sensitivity change 

Option: 1

0.75


Option: 2

0.25


Option: 3

0.50


Option: 4

0.10


Answers (1)

best_answer

  

 

 I_s = \frac{\alpha }{I} \Rightarrow V_s = \frac{\alpha }{V} = \frac{\alpha }{IR}= I_s/R

I' _s = I_s + (50/100 ) I_s = 3/2 I_s

V'_s = \frac{3/2 I_s}{2R}= 3/4 V_s = 0.75 V_s

 

 

 

 

Posted by

Sayak

View full answer

JEE Main high-scoring chapters and topics

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