- #1
Frigus
- 337
- 160
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.
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