Find the $emf$ $(E)$ of the cell in $V$ required to light the bulb rated $4.5 \, W, 1.5 \, V$ fully,as shown in the circuit.

Two resistors $400 \, \Omega$ and $800 \, \Omega$ are connected in series across a $6 \, V$ battery. The potential difference measured by a voltmeter of $10 \, k \Omega$ connected across the $400 \, \Omega$ resistor is close to $.... \, V$.

Three resistors $1 \Omega, 2 \Omega$ and $3 \Omega$ are connected to form a triangle. Across the $3 \Omega$ resistor,a $3 \text{ V}$ battery is connected. The current through the $3 \Omega$ resistor is: (in $\text{ A}$)

$A$ cell of emf $6\, V$ and internal resistance $0.5\, \Omega$ is short-circuited. The current in the cell is ............... $A$.

The figure shows a network of resistances and cells in which the point $A$ is earthed. The point which has the least potential is

Resistances $R_1$ and $R_2$,each $60\,\Omega$,are connected in series as shown in the figure. The potential difference between $A$ and $B$ is maintained at $120\,V$. What will be the reading of the voltmeter connected between points $C$ and $D$ if the resistance of the voltmeter is $120\,\Omega$?