Two cells of emf E₁ and E₂ and internal ...

CBSE Class 12 Physics 2023 Outside Delhi Set 2 Paper

Question : 13 of 13

Marks: +1, -0

Two cells of emf

E_1

and

E_2

and internal resistances

r_1

and

r_2

are connected in parallel, with their terminals of the same polarity connected together. Obtain an expression for the equivalent emf of the combination.

Solution: 👈: Video Solution

Terminal p.d of

1^{\text{st}}

cell,

V = E_1 - I_1 r_1

\therefore I_1 = \frac{E_1-V}{r_1}

Terminal p.d of

2^{\text{nd}}

cell,

V = E_2 - I_2 r_2

\therefore I_2 = \frac{E_2-V}{r_2}

\text{Now, } I = I_1 + I_2

\text{Or, } I = \frac{E_1-V}{r_1} + \frac{E_2-V}{r_2}

\text{Or, } I r_1 r_2 = r_2 (E_1-V) + r_1 (E_2-V)

\text{Or, } V (r_1+r_2) = E_1 r_2 + E_2 r_1 - I r_1 r_2

\text{Or, } V = \frac{E_1 r_2 + E_2 r_1}{r_1+r_2} - I \frac{r_2 r_1}{r_1+r_2}

Comparing with

V = E - I r

Equivalent emf

= E = \frac{E_1 r_2 + E_2 r_1}{r_1+r_2}

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