- #1

- 268

- 53

## Main Question or Discussion Point

My teacher proved it like this

I= ##\frac {E}{R+r}##

Where R is external resistance

And r is internal resistance

E is emf of the cell,

Now we can write it like this

I= ##\frac {E}{(√R-√r)²+2√R√r}##

if R=r then the denominator value will be minimum but i am unable to understand that why this is the case as for any value of R their will be some less value of r and vice versa which means current is not maximum when R=r.

Thanks.

I= ##\frac {E}{R+r}##

Where R is external resistance

And r is internal resistance

E is emf of the cell,

Now we can write it like this

I= ##\frac {E}{(√R-√r)²+2√R√r}##

if R=r then the denominator value will be minimum but i am unable to understand that why this is the case as for any value of R their will be some less value of r and vice versa which means current is not maximum when R=r.

Thanks.

Last edited: