The current in the given circuit is ................ $A$.

$A$ $16\, \Omega$ wire is bent to form a square loop. $A$ $9\, V$ supply having an internal resistance of $1\, \Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is $.......\, \times 10^{-1}\, V$.

In the given circuit,calculate the potential difference between points $P$ and $Q$ in $volt$.

The resistances of the four arms $P, Q, R$ and $S$ in a Wheatstone's bridge are $10 \, \Omega$, $30 \, \Omega$, $30 \, \Omega$ and $90 \, \Omega$, respectively. The e.m.f. and internal resistance of the cell are $7 \, V$ and $5 \, \Omega$ respectively. If the galvanometer resistance is $50 \, \Omega$, the current drawn from the cell will be ............... $A$.

The circuit shown is in a steady state. Now,when the switch is closed,the galvanometer shows no deflection. Then the correct relation is:

Four ammeters with identical internal resistances $r$ and a resistor of resistance $R$ are connected to a current source as shown in the figure. It is known that the reading of the ammeter $A_1$ is $I_1 = 3\ A$ and the reading of the ammeter $A_2$ is $I_2 = 5\ A$. Determine the ratio of the resistances $R/r$.