A galvanometer of resistance `(G) 10 Omega` and a series resistance `250 Omega` is used to convert it into a voltmeter of 2 V reading. The current passing through the voltmeter is nearly
A
0.0077 m A
B
0.077 m A
C
`0.77 m A
D
7.7 m A
Text Solution
Verified by Experts
The correct Answer is:
D
`I_g=(V)/(R+G)=(2)/(250+10)=7.7m A` .
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