A galvanometer has a coil of resistance 50 - and shows full deflection for 100 - A- The resistance to be added to the galvanometer to work as an ammeter of range 10 mA is

The correct option is B 0.5 Ω in parallel i _ 1 i _ 2 = R _ 2 R _ 1 = 100× 10 ^ 6 10× 10 ^ 3 = 50x 50+x #160; 50 #160; #160; (powe ...

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Standard XII

Physics

Voltmeter

A galvanomete...

Question

A galvanometer has a coil of resistance 50 $Ω$ and shows full deflection for 100 $Ω$ A. The resistance to be added to the galvanometer to work as an ammeter of range 10 mA is

0.5

Ω

in series

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0.5

Ω

in parallel

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Ω

in series

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Ω

in parallel

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Solution

The correct option is B 0.5 $Ω$ in parallel
$\frac{i_{1}}{i_{2}} = \frac{R_{2}}{R_{1}} = \frac{100 \times 10^{- 6}}{10 \times 10^{- 3}} = \frac{\frac{50 x}{50 + x}}{50}$ (power loss in both the cases will be same)

$= \frac{x}{50 + x} = \frac{1}{100}$
$100 x = 50 + x$
$99 x = 50$
$x = 0.5 Ω$

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A galvanometer has a coil of resistance 50 Ω and shows full deflection for 100 Ω A. The resistance to be added to the galvanometer to work as an ammeter of range 10 mA is

The correct option is B 0.5 Ω in paralleli1i2=R2R1=100×10−610×10−3=50x50+x50 (power loss in both the cases will be same) =x50+x=1100 100x=50+x 99x=50 x=0.5Ω

A galvanometer has a coil of resistance 50 $Ω$ and shows full deflection for 100 $Ω$ A. The resistance to be added to the galvanometer to work as an ammeter of range 10 mA is

The correct option is B 0.5 $Ω$ in parallel
$\frac{i_{1}}{i_{2}} = \frac{R_{2}}{R_{1}} = \frac{100 \times 10^{- 6}}{10 \times 10^{- 3}} = \frac{\frac{50 x}{50 + x}}{50}$ (power loss in both the cases will be same)

$= \frac{x}{50 + x} = \frac{1}{100}$
$100 x = 50 + x$
$99 x = 50$
$x = 0.5 Ω$