$A$ cell of $emf\;4\,V$ and internal resistance $0.5\,\Omega$ is connected to a $7.5\,\Omega$ external resistance. The terminal potential difference of the cell is $.....\,V$.

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 $\text{ V}$)

In the given circuit,if $V_{AB} = 4\ V$,then the value of resistance $X$ in ohms will be:

When a resistance of $2\,\Omega$ is connected across the terminals of a cell,the current is $0.5\,A$. When the resistance is increased to $5\,\Omega$,the current is $0.25\,A$. The internal resistance of the cell is ............. $\Omega$.

The reading of a high resistance voltmeter when a cell is connected across it is $2.2\, V$. When the terminals of the cell are also connected to a resistance of $5\,\Omega$,the voltmeter reading drops to $1.8\, V$. Find the internal resistance of the cell in $\Omega$.

The internal resistance of a cell of emf $4 \text{ V}$ is $0.1 \ \Omega$. It is connected to an external resistance of $3.9 \ \Omega$. The terminal voltage across the cell will be . . . . . . . (in $\text{ V}$)