JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 27)
To know the resistance G of a galvanometer by half deflection method, a battery of emf VE and resistance R is used to deflect the galvanometer by angle $$\theta $$. If a shunt of resistance S is needed to get half deflection then G, R and S are related by the equation :
2S (R + G) = RG
S (R + G) = RG
2S = G
2G = S
Explanation
When only galvanometer G is present with the resistance R,
_9th_April_Morning_Slot_en_27_1.png)
Here IG = $${{{V_E}} \over {R + G}}$$
When shunt of resistance S is connected parallel to galvanometer,
_9th_April_Morning_Slot_en_27_2.png)
Here I = $${{{V_E}} \over {R + {{GS} \over {G + S}}}}$$
As deflection is half, here current through galvanometer,
IG' = $${{{{\rm I}_G}} \over 2}$$
As both Galvanometer and shunt are parallel then potential are parallel then potential difference same.
$$ \therefore $$ IG' (G) = (I $$-$$ IG')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
Here IG = $${{{V_E}} \over {R + G}}$$
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,
IG' = $${{{{\rm I}_G}} \over 2}$$
As both Galvanometer and shunt are parallel then potential are parallel then potential difference same.
$$ \therefore $$ IG' (G) = (I $$-$$ IG')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
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