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Question
Two cells of emf E1 and E2 and internal resistances r1 and r2 are connected in parallel. Derive the expression for the (i) emf and (ii) internal resistance of a single equivalent cell which can replace this combination.
Solution
where
E1, E2 = Emf of two cells
r1, r2 = Internal resistances of cell
I1, I2 = Current due to the two cells
Terminal potential difference across the first cell is
V = E1 − I1r1
`=>I_1=(E_1-V)/r_1`
For the second cell, terminal potential difference will be equal to that across the first cell. So,
V = E2 − I2r2
`⇒I_2=(E_2−V)/r_2`
Let E be effective emf and r is effective internal resistance. Let I be the current flowing through the cell.
I=I1+I2
`⇒I=(E1−V)/r_1+(E2−V)/r_2`
`⇒Ir_1r_2=E_1r_2+E_1r_1−(r_1+r_2)V`
`=>V=(E_1r_2+E_2r_1)/(r_1+r_2)−(Ir_1r_2)/(r_1+r_2)`
Comparing the equation with V=E−Ir, we get
`E=(E_1r_2+E_2r_1)/(r_1+r_2)`
`r=(r_1r_2)/(r_1+r_2)`
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