The scale of a galvanometer is divided into $25$ equal divisions. The resistance of the galvanometer is $100 \,\Omega$. The current sensitivity of the galvanometer is $4 \times 10^{-4} \,A/div$. What resistance in $ohm$ must be connected in series to measure $2.5 \,V$?

  • A
    $100$
  • B
    $150$
  • C
    $250$
  • D
    $300$

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Similar Questions

$A$ galvanometer of resistance $40\,\Omega$ gives a deflection of $5\, \text{divisions}$ per $mA$. There are $50\, \text{divisions}$ on the scale. The maximum current that can pass through it when a shunt resistance of $2\,\Omega$ is connected is ................ $mA$.

In order to pass $10\,\%$ of the main current through a moving coil galvanometer of $99\,\Omega$,the resistance of the required shunt is ............ $\Omega$.

The resistance of a galvanometer is $50\,\Omega$ and the current required to give full-scale deflection is $100\,\mu A$. In order to convert it into an ammeter reading up to $10\,A$,it is necessary to put a resistance of

$STATEMENT-1$: The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil.
$STATEMENT-2$: Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.

In the adjoining circuit diagram,the readings of the ammeter and voltmeter are $2 \, A$ and $120 \, V$,respectively. If the value of $R$ is $75 \, \Omega$,then the voltmeter resistance will be (in $\Omega$):

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