$A$ resistor of resistance $R$ is connected between the terminals of a cell of $\operatorname{emf} E$ and internal resistance $r$. If $I$ is the current through the circuit,the terminal potential difference of the cell is given by:

In which of the following cells is the potential difference between the terminals of the cell greater than its $EMF$?

When two identical batteries of internal resistance $1 \Omega$ each are connected in series across a resistor $R$,the rate of heat produced in $R$ is $P_1$. When the same batteries are connected in parallel across $R$,the rate of heat produced is $P_2$. If $P_1 = 2.25 P_2$,then the value of $R$ is (in $\Omega$)

The emfs of three cells connected in parallel are $E_1=5 \ V, E_2=8 \ V$ and $E_3=10 \ V$ and their internal resistances are $r_1=1 \ \Omega, r_2=2 \ \Omega$ and $r_3=3 \ \Omega$,respectively. By changing $E_3$ to $E_{3N}$,the equivalent emf is doubled. Then $E_{3N}$ in $V$ is:

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,having the same $e.m.f.$ $E$,are connected in series through an external resistance $R$. The cells have internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ respectively. When the circuit is closed,the potential difference across the first cell is zero. The value of $R$ is