Question

Four cells each of emf $1.5\,V$ and internal resistance $2\, \Omega $ are connected in parallel. This combination sends a current to an external resistance $2\, \Omega $ . The value of current in external resistance is :

• A. 1.8 A
• B. 1.5 A
• C. 0.8 A
• D. 0.6 A

Answer 1

Answer: (D)

Net emf of combination of cells in parallel order = emf of each cell
$ =1.5V $ Net internal resistance is
$ {{r}_{in}}=\frac{r}{n}=\frac{2}{4}=0.5\Omega $ external resistance $ R=2\Omega $
So, current is given by
$ i=\frac{E}{R+{{r}_{in}}}=\frac{1.5}{2+0.5} $
$ =\frac{1.5}{2.5}=0.6A $