$A$ galvanometer coil has a resistance $80 \Omega$ and current for full-scale deflection is $10 \text{ mA}$. $A$ resistance of $920 \Omega$ is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is $0.2 \text{ V}$,the number of divisions on the scale is:

  • A
    $40$
  • B
    $46$
  • C
    $50$
  • D
    $92$

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Similar Questions

Consider two identical galvanometers and two identical resistors with resistance $R$. If the internal resistance of the galvanometers is $R_G < R/2$,which of the following configurations yield the maximum voltage range and maximum current range?

To determine the resistance $G$ of a galvanometer by the half-deflection method,a battery of $emf$ $V$ and a series resistance $R$ are used to produce a deflection $\theta$ in the galvanometer. If a shunt resistance $S$ is connected in parallel to the galvanometer to reduce the deflection to $\theta/2$,then $G, R$,and $S$ are related by the equation:

$A$ moving coil galvanometer has $48$ turns and the area of the coil is $4 \times 10^{-2} \, m^2$. If the magnetic field is $0.2 \, T$,then to increase the current sensitivity by $25\%$ without changing the area $(A)$ and the magnetic field $(B)$,the number of turns should become:

Explain the difficulties encountered when using a galvanometer directly as an ammeter.

$A$ milliammeter of resistance $40 \Omega$ has a range $0-30 \text{ mA}$. What will be the resistance used in series to convert it into a voltmeter of range $0-15 \text{ V}$ (in $\Omega$)?

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