$A$ current of $2.0 \, A$ passes through a cell of $e.m.f.$ $1.5 \, V$ having an internal resistance of $0.15 \, \Omega$. The potential difference measured,in $volts$,across the terminals of the cell will be:

Two sources of equal $emf$ $E$ are connected in series to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero,then:

Current flowing through the cell and potential difference between its two poles obtained by observations are given in the table below. The internal resistance of the cell is $r$.

Order	$V \ (V)$	$I \ (A)$
$1$	$1.0$	$0.08$
$2$	$0.5$	$0.18$
$3$	$0.8$	$0.12$

Find the $emf$ $(\varepsilon)$ of the cell used in the experiment. (in $V$)

$A$ battery has an $e.m.f.$ of $4\, V$ and an internal resistance $r$. When this battery is connected to an external resistance of $2\, \Omega$,a current of $1\, A$ flows in the circuit. How much current will flow if the terminals of the battery are connected directly?

In the circuit shown below,the current that flows from $a$ to $b$ when the switch $S$ is closed is ............... $A$.

The total power dissipated in Watts in the circuit shown here is