# Three cells of emf E, 2E and 5E having imyernal resistance r,2r and 3r. Variable reaistance R is shown in the figure. Find the expression for the current.

In given figure there is no variable resistance connected so i am assuming that it is connected in series with batteries.

The given condition is same as If two voltage sources are connected series with their emf in opposition and one battery is connected in series with with their emf in same direction.

If the batteries oppose one another, the total emf is less, since it is the algebraic sum of the individual emfs. When it is reversed, it produces an emf that opposes the other, and results in a difference between the two voltage sources.

So total emf = E+5E-2E

As all resistances whether internal or variable resistance are in series so total resistance

R

_{net}= r +3r+ 2r +R

So expression for current will be $I=\frac{E+5E-2E}{r+3r+2r+R}=\frac{4E}{6r+R}$

Regards

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