In the given diagram,an ideal voltmeter $V$ reads $1.45\,V$. Then the relation between $r_1$ and $r_2$ is:

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 $\text{ V}$)

Two cells,each of $e.m.f.$ $E$ and internal resistance $r$,are connected in parallel across an external resistor $R$. The maximum energy delivered to the resistor occurs when:

Two sources of equal $emf$ $(\varepsilon)$ 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:

$n$ rows,each containing $m$ cells in series,are connected in parallel. The maximum current is drawn from this combination through an external resistance of $3\, \Omega$. If the total number of cells used is $24$ and the internal resistance of each cell is $0.5\, \Omega$,then $m$ and $n$ are respectively:

$(a)$ Six lead-acid type of secondary cells each of $emf$ $2.0 \;V$ and internal resistance $0.015 \;\Omega$ are joined in series to provide a supply to a resistance of $8.5\; \Omega$. What are the current drawn from the supply and its terminal voltage?
$(b)$ $A$ secondary cell after long use has an $emf$ of $1.9 \;V$ and a large internal resistance of $380\; \Omega$. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?