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# The resistance of an ammeter is $13\Omega$ and its scale is graduated for current up to 100 A. After an additional shunt has been connected to this ammeter, it becomes possible to measure currents up to 750A by this meter. The value of shunt resistance isA.$2\Omega$B.$0.2\Omega$C.$2k\Omega$D.$20\Omega$

Last updated date: 13th Jun 2024
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Hint: A shunt is a resistance of very low value added across the terminals of ammeter whose range is to be extended. So in the above problem the shunt resistance can be calculated by comparing potential difference across the ammeter and that of shunt resistor.
Formula used:
$V=IR$

Shunt resistance is connected in parallel to increase the range of ammeter from 100A to750 A so, let us assume a shunt resistance ${{R}_{s}}$ is connected in parallel to A-B (ammeter in which resistance is shown outside for better understanding).
By adding shunt in parallel the maximum current ammeter can handle is still 100 A which is denoted by ${{I}_{S}}$ and the remaining current, will be equal to,
$\Rightarrow I-{{I}_{S}}$
Now,
$\Rightarrow I=750A$
$\Rightarrow {{I}_{S}}=100A$
Now, we know that potential across A-B will be equivalent to potential across shunt resistance so,
$\Rightarrow {{I}_{S}}R=(I-{{I}_{S}}){{R}_{S}}$
$\Rightarrow 100\times 13=(750-100){{R}_{S}}$
$\Rightarrow {{R}_{S}}=\dfrac{100\times 13}{650}=\dfrac{1300}{650}$
$\Rightarrow {{R}_{S}}=2\Omega$
$\therefore$ The required shunt resistance to be added in parallel to the ammeter to extend its range from 100 A to 750 A will be $2\Omega$

Hence, option (A) is correct.