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Question

A galvanometer gives full scale deflection of 1 volt when acting like a voltmeter when connected in series with 2 kΩ resistance. The same galvanometer gives 500 mA full scale deflection when acting like a ammeter when connected with shunt resistance of value 0.2 Ω in parallel. Find out the resistance of galvanometer (in Ω).

A
108
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B
222
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C
250
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D
1.5
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Solution

The correct option is B 222
Let internal resistance of galvanometer is G. Galvanometer is connected with R=2 kΩ in series combination
So. V=(R+G)Ig
Ig=V(R+G)
here, V=1 volt so,Ig=1(2000+G).......(1)
Now, an ammeter is connected in parallel with a resistance of 0.2 Ω
So, Ig×G=(500 mAIg)×0.2 Ω
Ig×G=(0.5AIg)×0.2
(0.2+G)×Ig=0.1
(0.2+G)×1(2000+G)=0.1.....[from eq. (1)]
(0.2+G)=200+0.1G
0.9G=199.8G=222 Ω
Hence, internal resistance of galvanometer is 222 Ω

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