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  4. 4 cells each of emf 2 V and internal resistance of 1Ω are connected in parallel to a load resistor of 2 Ω . Then the current through the load resistor is

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Q. 4 cells each of emf 2 V and internal resistance of $ 1\Omega$ are connected in parallel to a load resistor of $ 2 \Omega$ .
Then the current through the load resistor is

KEAMKEAM 2006Current Electricity

Solution:

The combined resistance of combination is
$ R' = R + \frac{r}{4} = 2 + \frac{1}{4} = \frac{9}{4} $
$ V = 2volt $
$ \therefore V = iR $
$ \Rightarrow i = \frac{V}{R} = \frac{2 \times 4}{9} = \frac{8}{9} $
$ = 0.888 A $

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