Two sources of equal emfs $E$ are connected in series. This combination is connected to an external resistance $R$. The internal resistances of the two sources are $r_{1}$ and $r_{2}$ $(r_{1} > r_{2})$. If the potential difference across the source of internal resistance $r_{1}$ is zero,then the value of $R$ will be:

The electromotive force of a primary cell is $2\,V$. When it is short-circuited,it gives a current of $4\,A$. Its internal resistance in $\Omega$ is:

Two cells $P$ and $Q$ each of emf $2.16 \text{ V}$ are connected in series with a resistor of $19.6 \text{ } \Omega$. An ideal voltmeter reads $2 \text{ V}$ when connected across the cell $P$ and $1.92 \text{ V}$ when connected across the cell $Q$. The ratio of the internal resistances of the cell $P$ and $Q$ is

$12$ cells,each having the same $emf$ $E$ and internal resistance $r$,are connected in series,but some cells are wrongly connected. This arrangement is connected in series with an ammeter and two additional cells (each of $emf$ $E$ and internal resistance $r$). The current is $3 \, A$ when the cells and the battery aid each other,and it is $2 \, A$ when they oppose each other. The number of cells wrongly connected is:

In an electric circuit,a cell of certain $emf$ provides a potential difference of $1.25\, V$ across a load resistance of $5\, \Omega$. However,it provides a potential difference of $1\, V$ across a load resistance of $2\, \Omega$. The $emf$ of the cell is given by $\frac{x}{10}\, V$. Then the value of $x$ is ..... .

A circuit consists of three batteries of emf $E_{1} = 1 \text{ V}$, $E_{2} = 2 \text{ V}$, and $E_{3} = 3 \text{ V}$ and internal resistances $1 \Omega$, $2 \Omega$, and $1 \Omega$ respectively, which are connected in parallel as shown in the figure. The potential difference between points $P$ and $Q$ is: (in $\text{ V}$)