Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
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:
Let total current be I
Given:
current through galvanometer = 5% of I
$$I_g=0.05I$$
So current through shunt:
$$I_s=0.95I$$
step 1: use parallel condition
Voltage across galvanometer = voltage across shunt
$$I_g\cdot G=I_s\cdot R_s$$
$$0.05I\cdot G=0.95I\cdot R_s$$
Cancel I:
$$0.05G=0.95R_s$$
$$R_s=\frac{0.05}{0.95}G=\frac{1}{19}G$$
step 2: equivalent resistance of ammeter
Galvanometer and shunt are in parallel:
$$R_A=\frac{G\cdot R_s}{G+R_s}$$
$$=\frac{G\cdot\frac{G}{19}}{G+\frac{G}{19}}=\frac{G^2/19}{G(1+1/19)}=\frac{G/19}{20/19}=\frac{G}{20}$$Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.