Physics, asked by asmaul3705, 1 year ago

A galvanometer of resistance, g, is shunted by a resistance s ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is

Answers

Answered by MidA
73

s ohm is connected in parallel as shunt to the galvanometer. so, let R is the resistance introduced in series with galvanometer.

final resistance = intial resistance (g)

so,

R + gs/(g+s) = g

=> R = g { 1 - s/(g+s)}

= g { (g + s - s)/(g +s)}

= g^2 / (g+s)................... ans...

Answered by VineetaGara
14

Answer: The value of the resistance which is parallel with the galvanometer is R=(G^2)÷(G+S)

Explanation:

The resistance of the galvanometer is G

The galvanometer is shunted by a resistance which is parallely connected to the galvanometer. The resistance of the shunt is S.

To put the main current unchanged a resistance is need to be connected to the galvanometer is series. Let assume the resistance of this resistor is R.

So,

=>{GS÷(S+G)}+R=G

=>R=G-{GS÷(S+G)

=>R=(G^2+GS-GS)÷(G+S)

=>R=(G^2)÷(G+S)

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