ig=10−4A
i=10×10−3A
RV=20kΩ
igG+igRV=5
igG=(i−ig)RA
RA=i−igigG=(i−ig)igig(5−igRV)
RA=i−ig5−igRV≈300 Ω
JEE Main 2019 — Physics Electromagnetism
A moving coil galvanometer allows a full scale current of 10−4A . A series resistance of 2×104Ω is required to convert the galvanometer into a voltmeter of range 0−5V . Therefore, the value of shunt resistance required to convert the above galvanometer into an ammeter of range 0−10mA is:
Held on 10 Apr 2019 · Verified 6 Jul 2026.
100Ω
200Ω
300Ω
10Ω
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