$A$ galvanometer with $30$ divisions has a current sensitivity of $20 \, \mu A/division$. Its resistance is $25 \, \Omega$. How will you convert it into an ammeter to measure $1 \, A$ current?

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

$A$ galvanometer has $36\,\Omega$ resistance. If a $4\,\Omega$ shunt is added to this,the fraction of current that passes through the shunt is

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 $M_1$.

An ideal ammeter and an ideal voltmeter have resistances of . . . . . . $\Omega$ and . . . . . . $\Omega$ respectively.

$A$ voltmeter essentially consists of