Question 339: Current Electricity

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 I
0
(I
0
> I
g
) by connecting a
shunt resistance R
A
to it and (ii) into a voltmeter of range 0 to V (V = GI
0
) by connecting a series resistance
R
V
to it. Then,

Accepted Answer

Galvanometer is converted into ammeter of range 0 to I
0
:
(I
o
- I
g
)R
A
= I
g
G
$$ \Rightarrow $$ $${R_A} = {{{I_g}G} \over {\left( {{I_0} - {I_g}} ight)}}$$............(1)
Galvanometer is converted into voltmeter of range 0 to V :
V = I
g
(R
V
+ G)
Also given, V = GI
0
$$ 	herefore $$ GI
0
= I
g
(R
V
+ G)
$$ \Rightarrow $$ $${R_V} = {{G\left( {{I_0} - {I_g}} ight)} \over {{I_g}}}$$ .........(2)
From equation (1) and (2)
$${R_A}{R_V} = {G^2}$$
$${{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} ight)^2}$$

Answer

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

Answer

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

Answer

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

Answer

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

Current Electricity

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