JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 7)
A moving coil galvanometer, having a resistance G, produces full scale deflection when a current Ig flows
through it. This galvanometer can be converted into (i) an ammeter of range 0 to I0(I0 > Ig) by connecting a
shunt resistance RA to it and (ii) into a voltmeter of range 0 to V (V = GI0) by connecting a series resistance
RV to it. Then,
$${R_A}{R_V} = {G^2}$$ and $${{{R_A}} \over {{R_V}}} = {{{I_g}} \over {\left( {{I_0} - {I_g}} \right)}}$$
$${R_A}{R_V} = {G^2}\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)$$ and $${{{R_A}} \over {{R_V}}} = {\left( {{{{I_0} - {I_g}} \over {{I_g}}}} \right)^2}$$
$${R_A}{R_V} = {G^2}\left( {{{{I_0} - {I_g}} \over {{I_g}}}} \right)$$ and $${{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}$$
$${R_A}{R_V} = {G^2}$$ and $${{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}$$
Explanation
Galvanometer is converted into ammeter of range 0 to I0 :
_12th_April_Evening_Slot_en_7_1.png)
(Io - Ig)RA = IgG
$$ \Rightarrow $$ $${R_A} = {{{I_g}G} \over {\left( {{I_0} - {I_g}} \right)}}$$............(1)
Galvanometer is converted into voltmeter of range 0 to V :_12th_April_Evening_Slot_en_7_2.png)
V = Ig(RV + G)
Also given, V = GI0
$$ \therefore $$ GI0 = Ig(RV + G)
$$ \Rightarrow $$ $${R_V} = {{G\left( {{I_0} - {I_g}} \right)} \over {{I_g}}}$$ .........(2)
From equation (1) and (2)
$${R_A}{R_V} = {G^2}$$
$${{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}$$
(Io - Ig)RA = IgG
$$ \Rightarrow $$ $${R_A} = {{{I_g}G} \over {\left( {{I_0} - {I_g}} \right)}}$$............(1)
Galvanometer is converted into voltmeter of range 0 to V :
V = Ig(RV + G)
Also given, V = GI0
$$ \therefore $$ GI0 = Ig(RV + G)
$$ \Rightarrow $$ $${R_V} = {{G\left( {{I_0} - {I_g}} \right)} \over {{I_g}}}$$ .........(2)
From equation (1) and (2)
$${R_A}{R_V} = {G^2}$$
$${{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}$$
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