Two batteries each of emf `epsilon` and internal resistance r, are connected in parallel. If we take current from this combination in a external resistance R, then for what value of R maximum power will beobtained? What will be this power?

A cell of emf epsilon and internal resistance r is charged by a current I, then

N batteries each of emf E and internal resistance r are first connected in series and then in parallel to an external resistance R. if current through R is same in both cases then.

Two identical cells each of emf epsilon , having negligible internal reistance r, are connercted in parallel with each other across an external resistance R. What is the current through this resistance.

Two batteris of emf epsilon_(1) and epsilon_(2) wit respective internal resistance r_(1) and r_(2) are connected in parallel. Now

An total resistance R is connected to a cell of internal resistance r the maximum current flows in the external resistance, when

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

Two cells of equal e.m.f. E b ut of difference internal resistance r_(1) and r_(2) are connected in parallel and the combination is connected with an external resistance R to obtain maximum power in R. find the value of R

Two batteries each of E and internal resistance r are connected turen by turn in series and in parallel, and areused to find current in an external resistance R . If the current in series is equal to that in parallel, the internal resistance of each battery is

Two cells , each of emf E and internal resistance r , are connected in parallel across a resistor R . The power delivered to the resistor is maximum if R is equal to