In the following Wheatstone bridge,$P/Q = R/S$. If key $K$ is closed,then the galvanometer will show deflection:

If the potential difference between points $B$ and $D$ is zero in the given circuit, the value of $x$ is $1/n \text{ } \Omega$. The value $n$ is

$A$ $16\, \Omega$ 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 circuit shown below,the ammeter reading is zero. Then the value of the resistance $R$ is (in $\Omega$)

As shown in the figure,a $5\,V$ battery with $2\,\Omega$ internal resistance and a $2\,V$ battery with $1\,\Omega$ internal resistance are connected to a $10\,\Omega$ resistor. The current flowing through the $10\,\Omega$ resistor is:

The value of current through the $5 \Omega$ resistor of the given circuit is