What is the minimum number of cells each of emf 10 V and internal resistance `1 Omega` to pass a current of 10 A through a resistance R of `3 Omega` ? What is the discharging power of R?

Explanation

By arraging n number of cells in each of the m numbers of rows we have,
`I=(mn e)/(mR+nr)`
To use the minimum number of cells, we require to fulfill the condition of maximum current. In this case,
`mR+nr or,m.3=n.1 or,n=3m`
Again , maximum current,
`I_0=(mn e)/(2mR)=(n e)/(2R)or,10=(ntimes10)/(2times3)or,n=6`
`thereforem=n/3=6/3=2`
So,12 cells need to be arranged in 2 rows and each row will have 6 cells.
Discharging power of `R,P=I_0^2R=(10)^2times3=300W` [See the chapter 'Electrical power and energy']