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,

Question

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,

TuteeHUB · Accepted Answer

${R_A}{R_V} = {G^2}\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right){\rm{\;and\;}}\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_0} - {I_g}}}{{{I_g}}}} \right)^2}$

TuteeHUB · Answer

${R_A}{R_V} = {G^2}\left( {\frac{{{I_0} - {I_g}}}{{{I_g}}}} \right){\rm{\;and\;}}\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_g}}}{{\left( {{I_0} - {I_g}} \right)}}} \right)^2}$

TuteeHUB · Answer

${R_A}{R_V} = {G^2}{\rm{\;and\;}}\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right)^2}$

TuteeHUB · Answer

${R_A}{R_V} = {G^2}{\rm{\;and\;}}\frac{{{R_A}}}{{{R_V}}} = \frac{{{I_g}}}{{\left( {{I_0} - {I_g}} \right)}}$

1.	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,
A.	\({R_A}{R_V} = {G^2}\left( {\frac{{{I_0} - {I_g}}}{{{I_g}}}} \right){\rm{\;and\;}}\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_g}}}{{\left( {{I_0} - {I_g}} \right)}}} \right)^2}\)
B.	\({R_A}{R_V} = {G^2}{\rm{\;and\;}}\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right)^2}\)
C.	\({R_A}{R_V} = {G^2}\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right){\rm{\;and\;}}\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_0} - {I_g}}}{{{I_g}}}} \right)^2}\)
D.	\({R_A}{R_V} = {G^2}{\rm{\;and\;}}\frac{{{R_A}}}{{{R_V}}} = \frac{{{I_g}}}{{\left( {{I_0} - {I_g}} \right)}}\)
Answer» C. \({R_A}{R_V} = {G^2}\left( {\frac{{{I_g}}}{{{I_0} - {I_g}}}} \right){\rm{\;and\;}}\frac{{{R_A}}}{{{R_V}}} = {\left( {\frac{{{I_0} - {I_g}}}{{{I_g}}}} \right)^2}\)

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