$A$ wire of resistance $R$ is bent in the form of a circular loop. Two points on the circle separated by a quarter circumference are connected to a battery of emf $E$ and negligible internal resistance. The heat generated in the wire per second is

One end each of a resistance $r$,a capacitor $C$,and a resistance $2r$ are connected together. The other ends are respectively connected to the positive terminals of batteries $P$,$Q$,and $R$ having emfs $E$,$E$,and $2E$. The negative terminals of the batteries are then connected together. In this circuit,with steady current,the potential drop across the capacitor is:

All wires have the same resistance $r$. The equivalent resistance between $A$ and $B$ is $R$. Now,if the keys $K_1$ and $K_2$ are closed,what will be the new equivalent resistance?

$A$ current of $2\,A$ flows through a $2\,\Omega$ resistor when connected across a battery. The same battery supplies a current of $0.5\,A$ when connected across a $9\,\Omega$ resistor. The internal resistance of the battery is:

The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated $220\,V$ and $100\,W$ is connected to a $(220 \times 0.8)\,V$ source,then the actual power consumed would be:

In the circuit shown in the figure,all cells are ideal. The current through the $2 \, \Omega$ resistor is ............ $A$.