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  • To find the resistance of a galvanometer by the half deflection method the following circuit is used with resistances <img alt="R_{1}=9970\Omega" src="https://learn.careers360.com/latex-image/

To find the resistance of a galvanometer by the half deflection method the following circuit is used with resistances R_{1}=9970\Omega,  R_{2}=30\Omega and R_{3}=0\Omega. The deflection in the galvanometer is d. With R_{3}=107\Omega the deflection changed to \frac{d}{2}. The galvanometer resistance is approximately : 

 

Option: 1

107\Omega
 


Option: 2

137\Omega


Option: 3

\frac{107}{2}\Omega

 


Option: 4

77\Omega


Answers (1)

best_answer

 

 

 

 

 

At initial condition, R3 = 0

Equivalent resistance of R2 & R3 = 30  Ω

When R3 =107 Ω & R2 = 30 Ω, then equivalent resistance should be 30/2

= 15 Ω

this will be when equivalent resistance , R(g) , R3 will be parallel to R2, we get resistance of 15 Ω

Let Rg - R3 = R(eq)

1/R2 + 1/R(eq) = 1/30 + 1/30 = 1/15

R(eq) = 30 Ω

∴ R3- Rg = 30

∴ 107 - Rg = 30

∴ Rg = 77 Ω

 

Posted by

Gaurav

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