How are internal resistance of individual cells related to internal resistance of the battery in case of series connection ?

Internal resistance of a cell depends on

For maximum power from battery the internal resistance of battery r is

In the series combination of n cells each cell having emf epsilon and internal resistance r. If three cells are wrongly connected, then total emf and internal resistance of this combination will be

A 6 V cell with 0.5Omega internal resistance , a 10 v cell with 1Omega internal resistance and a 12Omega external resistance are connected in parallel . The current (in amper) through the 10 V cell is

Write any two factors on which internal resistance of a cell depends . The reading on a high resistance voltmeter , when a cell is connected across it , is 2.0 V . When the terminals of the cell are also connected to a resistance of 3 Omega as shown in the circuit , the voltmeter reading drops to 1.5 V . Find the internal resistance of the cell.

n identical cells each of e.m.f. E and internal resistance r are connected in series. An external resistance R is connected in series to this combination. The current through R is

Four cells, each of emf E and internal resistance r, are connected in series across an external resistance R. By mistake one of the cells is connected in reverse. Then, the current in the external circuit is

There are n cells, each of emf E and internal resistance r, connected in series with an external resistance R. One of the cells is wrongly connected, so that it sends current in the opposite direction. The current flowing in the circuit 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

A 10 V battery with internal resistance 1 Omega and a 15 V battery with internal resistance 0.6 Omega are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to: