First slide

Moving coil Galvanometer

Question

A galvanometer of 50 ohm resistance has 25 divisions 'A current of 4x 10^-4 ampere gives a deflection of one division. To convert this galvanometer into a volt metre having a range of 25 volts, it should be connected with a resistance of

Easy

A

2500 $Ω$ as a shunt
B

245 $Ω$ as a shunt
C

2550 $Ω$ in series
D

2450 $Ω$ in series

Solution

Given that, G = 50 $Ω$
$i_{g} = 25 \times (4 \times 10^{- 4}) amp = 10^{- 2} amp$
To convert the galvanometer into voltmeter, high resistance should be connected in series with it. The value of high resistance fi is given by
$\begin{aligned} R = [\frac{V}{i_{g}} - G] = [\frac{25}{10^{- 2}} - 50] \\ = 2500 - 50 = 2450 Ω \end{aligned}$

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