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In the given circuit an external resistance 'R' is connected in parallel with 2 batteries or cells of e.m.f. E1 and E2 and internal resistance '0' and 'r2' respectively.Find i1,i2,i,Eequivalent, and r-equivalent for the given circuit.
Current distribution for the given circuit is considered as shown in figure .

Let us apply kirchoff's voltage law to the loop A-B-C-F-A ,

we get  i R + i2 r2 = E2

at junction F, we have i = i1 + i2

( i1 + i2 ) R + i2 r2 = E2   or  i1 R + i2 ( R + r2 ) = E2 ..................... (1)

Let us apply Kirrchoff's voltage law to the loop A-B-C-D-E-F-A, we get

i R =  i1 R + i2 R = E1 .......................(2)

From eqn.(1) and eqn.(2) , we get i2  = ( E2 - E1 ) / r2 ...................(3)

By substituting i2 in eqn.(2) and after simplification, we get , i1 = ( E1 / R ) - [ ( E2 - E1 ) / r2 ]  .....................(4)

From eqn.(3) and eqn.(4) , Total current i = i1 + i2 = ( E1 / R )

when two cells are connected in parallel, Equivalenet internal resistance  ( 1 / req ) = ( 1 / r1 ) + ( 1/ r2 )

Equivalent EMF is obtained from , ( Eeq / req ) = ( E1 / r1 ) + ( E2 / r2 )

Internal resistance r1 = 0 , makes it difficult to calculate equivalent internal resistance and equivalent EMF
Answered by Thiyagarajan K | 17 Oct, 2020, 08:13: PM

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