Consider a parallel combination of the cells shown in the figure. The potential difference between $B_1$ and $B_2$ is

$A$ galvanometer together with an unknown resistance in series is connected to two identical batteries each of $1.5\, V$. When the batteries are connected in series,the galvanometer records a current of $1\, A$,and when the batteries are in parallel,the current is $0.6\, A$. What is the internal resistance of the battery?

It is easier to start a car engine on a hot day than on a cold day. This is because the internal resistance of the car battery

Twelve cells,each having emf $E$ volts,are connected in series and are kept in a closed box. Some of these cells are wrongly connected with positive and negative terminals reversed. This $12$-cell battery is connected in series with an ammeter,an external resistance $R$ ohms,and a two-cell battery (two cells of the same type used earlier,connected perfectly in series). The current in the circuit when the $12$-cell battery and $2$-cell battery aid each other is $3 \text{ A}$,and it is $2 \text{ A}$ when they oppose each other. Then,the number of cells in the $12$-cell battery that are connected wrongly is:

Ten identical cells each emf $2 \, V$ and internal resistance $1 \, \Omega$ are connected in series with two cells wrongly connected. $A$ resistor of $10 \, \Omega$ is connected to the combination. What is the current through the resistor (in $ \, A$)?

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: