The e.m.f. of a cell is $E \text{ volts}$ and its internal resistance is $r \text{ ohm}$. The resistance in the external circuit is also $r \text{ ohm}$. The potential difference across the cell will be

$A$ constant voltage of $50 \ V$ is maintained between the points $A$ and $B$ of the circuit shown in the figure. The current through the branch $CD$ of the circuit is (in $A$)

Resistors of $1\, \Omega$, $2\, \Omega$, and $3\, \Omega$ are connected in the form of a triangle. If a $1.5\, V$ cell of negligible internal resistance is connected across the $3\, \Omega$ resistor, the current flowing through this resistance will be ................ $A$.

$A$ battery of $6 \text{ V}$ is connected to the ends of a uniform wire $3 \text{ m}$ long and of resistance $100 \Omega$. The potential difference between two points $50 \text{ cm}$ apart on the wire is: (in $\text{ V}$)

The equivalent resistance of the infinite network given below is:

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 $i_1: i_2=$