The scale of a galvanometer is divided into 150 equal divisions. The galvanometer has the current sensitivity of 10 divisions per mA and the voltage sensitivity of 2 divisions per mV. How the galvanometer be designed to read i) 6A, per division and ii) 1V, per division ?

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$40 μ Ω$ is connected in parallel, $1000 Ω$ is connected in series

$8.3 \times 10^{- 5} Ω$ is connected in parallel, $9995 Ω$ is connected in series

$8.3 μ Ω$ is connected in parallel, $9995 Ω$ is connected in series

$20 μ Ω$ is connected in parallel, $9995 Ω$ is connected in series

answer is D.

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Detailed Solution

$N = 150; i_{g} = 0.1 mA / division$

$V_{g} = 0.5 mV / division ; G = \frac{V_{g}}{i_{g}} = 5 Ω$

(i) shunt required to read i = 6A/division is $S = \frac{{Gi}_{g}}{i - i_{g}} = 8.3 \times 10^{- 5} Ω (in parallel)$

(ii) Resistance required to read the voltage $V = 1 V / division is = G ((V / V_{g}) - 1) = 9995 Ω$ (in series)

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