Answer
414.9k+ views
Hint: Shunt resistance is a resistor with such a type of resistor having a very low resistance value is called shunt resistance. The shunt resistor is usually constructed of a substance with a resistance coefficient of low temperature. It is related to the ammeter, whose range is to be expanded, in parallel.
Formula used:
For solving this question, we will be using the formula for the Shunt resistance, i.e.,
$S=\dfrac{{{I}_{g}}G}{I-{{I}_{g}}}$
Complete answer:
Let us take a look at all the given parameters,
G=\[900\Omega \]
Let the main current be I
Now, current through galvanometer will be 10% of the main current,
\[\Rightarrow {{I}_{g}}=0.1I\]
So, current through shunt will be 90% of the main current,
\[\Rightarrow {{I}_{S}}=0.9I\]
Now, applying the formula for the resistance of the shunt, that we discussed above
$S=\dfrac{{{I}_{g}}G}{I-{{I}_{g}}}$
Using the given parameters in the formula for the resistance of shunt
$\Rightarrow S=\dfrac{0.1I\times 900}{I-0.1I}$
$\Rightarrow S=100\Omega $
So, in order to send only 10% of the main current through the galvanometer.
The resistance of the required shunt will be Option – B, i.e., $100\Omega $.
Note:
The galvanometer is an electromechanical device used to detect and detect electrical current. The galvanometer operates as an actuator, producing a rotary deflection ("pointer") in response to the electrical current flowing through the coil in a constant magnetic field. The first galvanometer was reported by Johann Schweigger at the University of Halle on 16 September 1820. André-Marie Ampère has also contributed to its growth. Early designs improved the influence of the magnetic field produced by the current by using several wire turns.
Formula used:
For solving this question, we will be using the formula for the Shunt resistance, i.e.,
$S=\dfrac{{{I}_{g}}G}{I-{{I}_{g}}}$
Complete answer:
Let us take a look at all the given parameters,
G=\[900\Omega \]
Let the main current be I
Now, current through galvanometer will be 10% of the main current,
\[\Rightarrow {{I}_{g}}=0.1I\]
So, current through shunt will be 90% of the main current,
\[\Rightarrow {{I}_{S}}=0.9I\]
Now, applying the formula for the resistance of the shunt, that we discussed above
$S=\dfrac{{{I}_{g}}G}{I-{{I}_{g}}}$
Using the given parameters in the formula for the resistance of shunt
$\Rightarrow S=\dfrac{0.1I\times 900}{I-0.1I}$
$\Rightarrow S=100\Omega $
So, in order to send only 10% of the main current through the galvanometer.
The resistance of the required shunt will be Option – B, i.e., $100\Omega $.
Note:
The galvanometer is an electromechanical device used to detect and detect electrical current. The galvanometer operates as an actuator, producing a rotary deflection ("pointer") in response to the electrical current flowing through the coil in a constant magnetic field. The first galvanometer was reported by Johann Schweigger at the University of Halle on 16 September 1820. André-Marie Ampère has also contributed to its growth. Early designs improved the influence of the magnetic field produced by the current by using several wire turns.
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