Two identical cells of emf 1.5V each joined in parallel supply energy to an external circuit consisting of two resistance of `7 Omega` each joined in parallel. A very high resistance voltmeter reads the terminal voltage of cells to be 1.4V.Calculate the internal resistance of each cell.

Explanation

The two cells are connected in parallel. So, the equivalent emf is 1.5V.
Now, the two resistors are connected in parallel.So, the equivalent resistance to
`1/R_(eq)=1/R+1/R=2/R`
`thereforeR_(eq)=R/2=7/2=3.5Omega`
The terminal voltage of the cells measured by the voltmeter is 1.4V.
The equivalent internal resistance of the combination of cells is
`r_(eq)=((E-V)/V)R`
`thereforer_(eq)=(1.5-1.4)/1.4times3.5=0.1/1.4times3.5=0.25Omega`
As the cells are connected in parallel,
so, `r_(eq)=(r')/2[r'`= internal resistance of each cell]
`thereforer'=2r_(eq)=2times0.25=0.5Omega`