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  • A moving coil galvanometer is converted into an ammeter reading upto 0.03 A by connecting a shunt of resistance 4r across it and into an ammeter reading upto 0.06 A when a shunt of resistance r is

A moving coil galvanometer is converted into an ammeter reading upto 0.03 A by connecting a shunt of resistance 4r across it and into an ammeter reading upto 0.06 A when a shunt of resistance r is connected across it. What is the maximum current which can be sent through this galvanometer, if no shunt is used?

Option: 1

0.01 A


Option: 2

0.02 A


Option: 3

0.03 A


Option: 4

0.04 A


Answers (1)

best_answer

For an ammeter, \mathrm{\frac{\mathrm{i}_{\mathrm{g}}}{\mathrm{i}}=\frac{\mathrm{S}}{\mathrm{G}+\mathrm{S}}}

\begin{aligned} & \Rightarrow \mathrm{i}_{\mathrm{g}} \mathrm{G}=\left(\mathrm{i}-\mathrm{i}_{\mathrm{g}}\right) \mathrm{S} \\ \\& \therefore \mathrm{i}_{\mathrm{g}} \mathrm{G}=\left(0.03-\mathrm{i}_{\mathrm{g}}\right) 4 \mathrm{r}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \text{(i)} \end{aligned}

and \mathrm{i}_{\mathrm{g}} \mathrm{G}=\left(0.06-\mathrm{i}_{\mathrm{g}}\right) \mathrm{r} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (\text{ii})

From eqs. (i) and (ii), we get

0.12-4 \mathrm{i}_{\mathrm{g}}=0.06-\mathrm{i}_{\mathrm{g}}

Maximum current through the galvanometer, \mathrm{i}_{\mathrm{g}}=0.02 \mathrm{~A}

Posted by

Devendra Khairwa

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