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Obtain an expression for the equivalent ...

Obtain an expression for the equivalent emf and internal resistance of two cells connected in parallel.

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Consider two cells of emf `epsilon_1` & `epsilon_2` and corresponding internal resistances `r_1, r_2` are connected in parallel. Since as much charge flows in as out, we have
`I=I_1+I_2`
Both the cells are at same potential since they are in parallel .
`V=epsilon_1 - I_1r_1` For first cell
`V=epsilon_2-I_2r_2` For second cell

So, `I_1=(epsilon_1-V)/r_1, I_1=(epsilon_2-V)/r_2`
`I=(epsilon_1-V)/r_1 + (epsilon_2-V)/r_2=epsilon_1/r_1+epsilon_2/r_2-V(1/r_1+1/r_2)`
`I=(epsilon_1 r_2+epsilon_2r_1)/(r_1r_2)-V((r_2+r_1)/(r_1r_2)) rArr I(r_1r_2)=epsilon_1r_1+epsilon_2r_2 - V(r_2+r_1)`
`rArr I((r_1r_2))/((r_1+r_2)) = (epsilon_1r_2+epsilon_2r_1)/((r_1+r_2))-V`
`therefore V=(epsilon_1r_2+epsilon_2r_1)/((r_1+r_2))-V((r_1r_2))/((r_1+r_2))`
`V=epsilon_"equivalent"-Ir_"equivalent"`
`therefore epsilon_(eq)=(epsilon_1r_2+epsilon_2r_1)/((r_1+r_2))` and `r_(eq)=((r_1r_1))/((r_1+r_2))` or `1/r_(eq)=1/r_1+1/r_2`
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