Answer
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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:
$V=IR$
Complete answer:
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.
Additional information:
The resistance and inductive reactance is very low so it cannot be connected in parallel because in parallel due to low resistance it can short the circuit and maximum current will pass through it, by which instrument might burn. For ideal ammeters, the impedance is zero by which voltage drop across the instrument will be zero.
Note:
Ammeters can be of different types depending on the construction and measurement of current to be needed. Based on construction some types are permanent moving coil ammeter, moving iron ammeter etc. and for measurement of current we have AC ammeter or DC ammeter.
Formula used:
$V=IR$
Complete answer:
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.
Additional information:
The resistance and inductive reactance is very low so it cannot be connected in parallel because in parallel due to low resistance it can short the circuit and maximum current will pass through it, by which instrument might burn. For ideal ammeters, the impedance is zero by which voltage drop across the instrument will be zero.
Note:
Ammeters can be of different types depending on the construction and measurement of current to be needed. Based on construction some types are permanent moving coil ammeter, moving iron ammeter etc. and for measurement of current we have AC ammeter or DC ammeter.
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