WH2o - W1 3420 3) at is desired to pass only 2% of the total current...

Question

Question

WH2o - W1 3420 3) at is desired to pass only 2% of the total current through a galvanometre resistence as r. How will u acchi

Answer 1

Answer #1

For this purpose connect a low resistance S in parallel with Galavanometer (G). Hence total current I is divided into G and I-Ig through S

In parallel combinaton Voltage is same

IgXG=(I-Ig)XS

G=resistance of galvanometer=98 $\Omega$

Ig=2% of I=2I/100=.02I

IgXG=(I-Ig)XS

S=IgXG/(I-Ig)=(0.02IX98)/(I-.02I)=2 $\Omega$

That is when we connect 2 $\Omega$ in parallel with galvanometer, 2% of current passes through it

Answer 2

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