Two meters of voltage range 20.0V and 30.0V have to be constructed with a galvanometer. The resistance connected in series with the galvanometer is 1680 Ω for the 20.0V range and 2930 Ω for the 30.0V range. The resistance of the galvanometer and the full -scale current are respectively.
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
320 Ω and 8 mA
b
70 Ω and 10 mA
c
820 Ω and 10 mA
d
820 Ω and 8 mA
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Given that v=20.0V for R1=1680Ωv=30.0V for R2=2930ΩLet coil resistance = GIf full scale current is IgIgG+R1=20⇒Ig(G+1680)=20…..(1) And Iσ(G+2930)=30…..(2)Dividing (1) & (2) G+1680G+2930=2030⇒3(G+1680)=2(G+2930)⇒G=5860−5040∴G=820ΩFrom (1)Ig=20820+1680=8×10−3A∴Ig=8mA