Two cell emf and internal resistance \varepsilon _{1}, r_{1} and \varepsilon _{2}, r_{2} are connected in parallel. Derive the expressions for the emf and internal resistance of a cell which can replace this combination.

 

Answers (1)
S safeer

Emf are \varepsilon _{1}and \varepsilon _{2}

Internal resistance are r _{1}andr _{2}

Here, I=I _{1}+I_{2}            ........(1)

Let V is the potential differnace between B1 and B2

Then V = \varepsilon _{1}-I_{1}r_{1}

Therefore , I_{1}= \frac{\varepsilon _{1}-V}{r_{1}}                ......(2)

Similarly. I_{2}= \frac{\varepsilon _{2}-V}{r_{2}}                   .......(3)

Putting equation (2) and (3) in (1)

I=\frac{\varepsilon _{1}-V}{r_{1}}+ \frac{\varepsilon _{2}-V}{r_{2}}

 I=\left ( \frac{\varepsilon _{1}}{r_{1}}+\frac{\varepsilon _{2} }{r_{2}} \right )-V\left ( \frac{1}{r_{1}}+\frac{1}{r_{2}} \right )

or V = \left ( \frac{\varepsilon _{1}r_{2}+\varepsilon _{2}r_{1}}{r_{1}+r_{2}} \right )- I\left ( \frac{r_{1}r_{2}}{r_{1}+r_{2}} \right )

If we replace this with a single cell it can be written as 

V = \varepsilon _{equivalent}-Ir_{eqivalent}

Then, 

\varepsilon _{eqivalent} = \left ( \frac{\varepsilon _{1}r_{2}+\varepsilon _{2}r_{1}}{r_{1}+r_{2}} \right )

r_{equivalent} =\left ( \frac{r_{1}r_{2}}{r_{1}+r_{2}} \right )

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