Two cells of emf E1 and E2 and internal resistance r1 and r2 are connected in parallel

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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Obtain the equivalent emf and internal resistance of a parallel combination of 2 cell of emf E1 and E2 and internal resistance R1 and R2.

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Text Solution

Solution : If the equivalent emf of the combination is `F_0` and its equivalent internal resistance is `r_0` then from the figure 1.105, <br> `r_.=(r_1r_2)/(r_1+r_2),l_1=E_1/r_1,l_2=E_2/r_2andl=E_0/r_0`<br> <img src="//d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_DMB_PHY_XII_P1_U02_C01_E06_016_S01.png" width="80%"> <br> `becausel=l_1+l_2,thereforeE_0/r_0=E_1/r_1+E_2/r_2`<br> or,`E_0=E_1r_0/r_1+E_2r_0/r_2=E_1r_2/(r_1+r_2)+E_2r_1/(r_1+r_2)`<br> `=1/(r_1+r_2)(E_1r_2+E_2r_1)`