First slide

Tangent Galvanometer

Question

The seave of a galvanometer of resistance 100 $Ω$ contains 25 divisions. It gives a deflection of 1 division on passing a current of 4 x 10^-4 A. The resistance in ohm to be added to it, so that it may become a voltmeter of range 2 .5 V is

Moderate

A

100
B

150
C

250
D

300

Solution

For full scale deflection , i_g = n k
$∴ i_{g} = 25 \times (4 \times 10^{- 4}) = 10^{- 2} A$
Now $R = \frac{V}{i_{g}} - G = \frac{2 \cdot 5}{10^{- 2}} - 100$
$= 250 - 100 = 150 Ω$

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