$A$ galvanometer has a resistance of $100 \Omega$ and a current of $10 \text{ mA}$ produces full-scale deflection in it. The resistance to be connected to it in series,to convert it into a voltmeter of range $50 \text{ V}$,is: (in $\Omega$)

  • A
    $3900$
  • B
    $4000$
  • C
    $4600$
  • D
    $4900$

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$A$ certain current passing through a galvanometer produces a deflection of $100$ divisions. When a shunt of $1 \ \Omega$ is connected, the deflection reduces to $1$ division. The galvanometer resistance is: (in $\Omega$)

Two moving coil meters $M_1$ and $M_2$ have the following particulars:
$R_1 = 10\,\Omega, N_1 = 30, A_1 = 3.6 \times 10^{-3}\, m^2, B_1 = 0.25\, T$
$R_2 = 14\,\Omega, N_2 = 42, A_2 = 1.8 \times 10^{-3}\, m^2, B_2 = 0.50\, T$
(The spring constants are identical for the two meters). Determine the ratio of voltage sensitivity of $M_2$ and $M_1$.

$A$ student is provided with a variable voltage source $V$,a test resistor $R_T=10\,\Omega$,two identical galvanometers $G_1$ and $G_2$,and two additional resistors,$R_1=10\,M\,\Omega$ and $R_2=0.001\,\Omega$. For conducting an experiment to verify Ohm's law,the most suitable circuit is:

When a shunt of $4\,\Omega$ is attached to a galvanometer,the deflection reduces to $\frac{1}{5}$th of its initial value. If an additional shunt of $4\,\Omega$ is attached in parallel to the first,what will be the new deflection?

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$A$ moving coil galvanometer has a resistance of $50\,\Omega$ and gives full-scale deflection for $10\,mA$. How could it be converted into an ammeter with a full-scale deflection for $1\,A$?

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