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$)?

The internal resistance of a cell depends on

Infinite number of cells having $emf$ and internal resistance $(E, r)$,$(\frac{E}{n}, \frac{r}{n})$,$(\frac{E}{n^2}, \frac{r}{n^2})$,$(\frac{E}{n^3}, \frac{r}{n^3})$... are connected in series across an external resistance of $\frac{nr}{n+1}$. The current flowing through the external resistor is:

$10$ cells,each of $emf$ $E$ and internal resistance $r$,are connected in series to a variable external resistance. The figure shows the variation of the terminal potential difference of their combination with the current drawn from the combination. The $emf$ of each cell is ................ $V$.

$A$ cell of emf $10 \text{ V}$ and internal resistance $3 \Omega$ is connected in parallel with another cell of emf $7 \text{ V}$ and internal resistance $\frac{3}{5} \Omega$,such that their positive terminals are joined together and their negative terminals are joined together. Their combined positive terminal is joined with the negative terminal,and their combined negative terminal is joined with the positive terminal of a third cell of emf $20 \text{ V}$ with internal resistance $2 \Omega$. The combination can be replaced by a battery of emf $E$ and internal resistance $r$. The values of $E$ and $r$ are respectively:

Two batteries,one of emf $18 \, V$ and internal resistance $2 \, \Omega$ and the other of emf $12 \, V$ and internal resistance $1 \, \Omega$,are connected in parallel as shown. The voltmeter $V$ will record a reading of ............. $V$.