In steady state, the potential difference across the capacitor is $10\, V$. Each resistance is of $3\, \Omega$. The cell is ideal. The $emf$ of the cell is ............. $V$.

$A$ uniform metallic wire having resistance $4 \ \Omega$ is bent to form a square loop $(ABCD)$ (see figure). $A$ resistance of $2 \ \Omega$ is connected between points $B$ and $D$ and a battery of $2 \ V$ is connected across points $A$ and $C$ as shown in the figure. Now the value of current $(I)$ is: (in $A$)

In the given circuit,the internal resistance of the cell is zero. If $i_1$ and $i_2$ are the readings of the ammeter when the key $(K)$ is opened and closed respectively,then $i_1: i_2=$

Two resistors of $6\,\Omega$ and $9\,\Omega$ are connected in series to a $120\,V$ source. The power consumed by the $6\,\Omega$ resistor is ........... $W$.

$50\,\Omega$ and $100\,\Omega$ resistors are connected in series. This combination is connected to a battery of $2.4\,V$. When a voltmeter of $100\,\Omega$ resistance is connected across the $100\,\Omega$ resistor,the reading of the voltmeter will be ............. $V$.

The galvanometer deflection,when key $K_1$ is closed but $K_2$ is open,equals $\theta_0$ (see figure). On closing $K_2$ also and adjusting $R_2$ to $5\,\Omega$,the deflection in the galvanometer becomes $\frac{\theta_0}{5}$. The resistance of the galvanometer is,then,given by [Neglect the internal resistance of battery]: .................. $\Omega$