Name: Conversion of a Galvanometer Into a Voltmeter
Uploaded: 2019-07-19
Duration: 7 min 18 s
Description: This video helps us understand the working of a galvanometer. It also teaches us how to convert it into a voltmeter of the desired range both theoretically a...

Question

A multirange current meter can be constructed by using a galvanometer circuit as shown in the figure. We want a current meter that can measure $10 mA, 100 mA and 1 A$ using a galvanometer of resistance $10 Ω$ and that prduces maximum deflection for current of $1 mA$ . Find $S_{1}, S_{2} and S_{3}$ that have to be used

Accepted Answer

We can measure the currents of magnitude $10 mA, 100 mA$ and $1 A$ by connecting ammeter $A$ and $B, C$ and $D$ respectively. So,

$For 10 mA \to I_{g} G = (I - I_{g}) (S_{1} + S_{2} + S_{3}) \dots (I)$

$For 100 mA \to I_{g} (G + S_{1}) = (I - I_{g}) (S_{2} + S_{3}) \dots (II)$

$For 1 A \to I_{g} (G + S_{1} + S_{2}) = (I - I_{g}) S_{3} \dots (III)$

$I_{g} = 1 mA = 10^{- 3} A and G = 10 Ω$

$10^{- 3} θ 10 = (10^{- 2} - 10^{- 3}) (S_{1} + S_{2} + S_{3}) [from (I)]$

$\Rightarrow 10 = (10 - 1) (S_{1} + S_{2} + S_{3})$

$\Rightarrow 10 = 9 (S_{1} + S_{2} + S_{3}) \dots (IV)$

$10^{- 3} (10 + S_{1}) = (10^{- 1} - 10^{- 3}) (S_{2} + S_{3}) [from (I I)]$

$\Rightarrow 10 + S_{1} = (100 - 1) (S_{2} + S_{3})$

$\Rightarrow 10 + S_{1} = 99 (S_{2} + S_{3}) \dots (V)$

$10^{- 3} (10 + S_{1} + S_{2}) = (1 - 10^{- 3}) (S_{3}) [from (III)]$

$\Rightarrow 10 + S_{1} + S_{2} = (1000 - 1) S_{3}$

$\Rightarrow 10 + S_{1} + S_{2} = 999 S_{3} \dots (V I)$

$10 + S_{1} = 99 (S_{3} + S_{2}) [from (V)]$

$10 = 9 (S_{3} + S_{2} + S_{1}) [from (V)]$

$\frac{10}{9} = S_{1} + S_{2} + S_{3} [from (IV)]$

$\frac{\frac{10}{99} + \frac{S_{1}}{99} = \frac{S_{2} + S_{3}}{-}}{\frac{10}{9} - \frac{10}{99} - \frac{S_{1}}{99} = S_{1}} [from (V)]$

$\frac{110 - 10}{99} = S_{1} + \frac{S_{1}}{99}$

$\frac{100}{99} = \frac{99 S_{1} + S_{1}}{99} \Rightarrow \frac{100}{99} = \frac{100}{99} S_{1}$

$S_{1} = 1 Ω$

$\frac{10}{9} = 1 + S_{2} + S_{3}$ from (VI)

Or $\frac{1}{9} = S_{2} + S_{3}$

$S_{3} = \frac{10}{999} + \frac{S_{1}}{999} + \frac{S_{2}}{999}$ from (VI)
$S_{3} = \frac{10}{999} + \frac{1}{999} + \frac{S_{2}}{999} [∵ S_{1} = 1 Ω]$

$\Rightarrow S_{3} = \frac{11}{999} + \frac{S_{2}}{999}$

$S_{2} = \frac{1}{9} - S_{3} [∵ \frac{1}{9} = S_{2} + S_{3} from above]$

$= \frac{1}{9} - \frac{11}{999} - \frac{S_{2}}{999}$ from (III)

$S_{2} + \frac{S_{2}}{999} = \frac{111 - 11}{999}$

$\frac{1000}{999} S_{2} = \frac{100}{999}$

$S_{2} = 0.1 Ω$

$S_{1} + S_{2} + S_{3} = \frac{10}{9}$ from (IV)

$\Rightarrow 1 + 0.1 + S_{3} = \frac{10}{9}$

$\Rightarrow S_{3} = \frac{10}{9} - \frac{11}{10} = \frac{100 - 99}{90} = \frac{1}{90}$

Here $S_{1} = 1 Ω, S_{2} = 0.1 Ω; S_{3} = 0.010 Ω$

Important Questions on Moving Charges and Magnetism

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A multirange current meter can be constructed by using a galvanometer circuit as shown in the figure. We want a current meter that can measure 10mA, 100mA and 1A using a galvanometer of resistance 10Ω and that prduces maximum deflection for current of 1mA. Find S1,S2 and S3 that have to be used

Important Questions on Moving Charges and Magnetism

Learn and Prepare for any exam you want !