Two cells are connected in opposition as shown.Cell `E_(1)` is of "8V" emf and `2Omega` internal resistance, the cell `E_(2)` is of "2V" emf and `4Omega` internal resistance.The terminal potential difference of cell `E_(2)` is __________ V.

The circuit in Fig. shows two cells connected in opposition to each other. Cell E_(1) is of emf 6V and internal resistance 2Omega, the cell E_(2) is of emf 4 V sand internal resistance 8omega . Find the potential difference between the points A and B .

Two cells e_(1) and e_(2) connected in opposition to each other as shown in figure. The cell of emf 9 V and internal resistance 3Omega the cell is of emf 7V and internal resistance 7Omega . The potential difference between the points A and B is

A cell of e.mf. E and internal resistance r is connected in series with an external resistance nr. Then, the ratio of the terminal potential difference to E.M.F.is

Two cells of emf E_(1) and E_(2) are joined in opposition (such that E_(1) gt E_(2) ) . If r_(1) be the internal resistances and R be the external resistance, then the terminal potential difference is

A cell E_1 of emf 6V and internal resistance 2 Omega is connected with another cell E_2 of emf 4V and internal resistance 8 Omega (as shown in the figure). The potential difference across points X and Y is :

In a connection of 10 dry cells each of emf E and internal resistance r in series, two cell is connected opposite. Net emf and net internal resistance will be: