
As potential drop across both branches would be the same, we get ISS=IgG
(10095I)S=(1005I)G
⇒S=19G
Now, resistance of ammeter will be
RA=S+GSG=1920G19G2
⇒RA=20G
JEE Main 2024 — Physics Electromagnetism
In an ammeter, 5 of the main current passes through the galvanometer. If resistance of the galvanometer is G, the resistance of ammeter will be:
Held on 1 Feb 2024 · Verified 6 Jul 2026.
20G
199G
199G
200G
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