QJEEM16Ph0334: JEE Main 2016 Physics - Class 12 JEEM16Ph0334 Solution: To know the resistance G of a galvanometer by half deflection…

When only galvanometer G is present with the resistance R,



Here I_G = ${{{V_E}} \over {R + G}}$

When shunt of resistance S is connected parallel to galvanometer,



Here I = ${{{V_E}} \over {R + {{GS} \over {G + S}}}}$

As deflection is half, here current through galvanometer,

I_G' = ${{{{\rm I}_G}} \over 2}$

As both Galvanometer and shunt are parallel then potential are parallel then potential difference same.

$\therefore$    I_G' (G) = (I $-$ I_G^')S

$\Rightarrow$    I'_G (G + S) = IS

$\Rightarrow$     ${{{{\rm I}_G}} \over 2}$ = ${{{\rm I}S} \over {G + S}}$

$\Rightarrow$     ${{{V_E}} \over {2\left( {R + G} \right)}}$ = ${{{V_E}} \over {R + {{GS} \over {G + S}}}}$ $\times$ ${S \over {\left( {G + S} \right)}}$

$\Rightarrow$     ${1 \over {2\left( {R + G} \right)}}$ = ${{G + S} \over {R(G + S) + GS}}$ $\times$ ${S \over {\left( {G + S} \right)}}$

$\Rightarrow$    RG + RS + GS = 2RS + 2GS

$\Rightarrow$    RG = RS + GS

$\Rightarrow$     S(R + G) = RG