The maximum permissible current through a galvanometer G of internal resistance 10\(\Omega\) is 0.05A. A resistance R is used to convert G into a voltmeter with a maximum reading of 100V. Find the value of R and how it is connected to G

a

20,000 ohms in parallel

b

19,990 ohms in series

c

1990 ohms in series

d

100 ohms in series

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c

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Discussions (13)

olyezema
4 years ago
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brainiac140
5 years ago
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FGGCI161984
7 years ago

I don't understand how it is solve
๐Ÿ˜•๐Ÿ˜•.pls help me out.

Godswilldan
2 months ago

love that

Habibolly
2 months ago

100 รท 0.05 =2000
2000 - 10 = 1990

VictorAgulonu
1 year ago

A multiplier (high serie resistance (

right

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GloryAbiola
6 years ago

someone should please explain it,i don't understand it

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