Question

Four cells each of emf $1.5V$ 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
$i=\frac{E}{R+r_{in}}=\frac{1.5}{2+0.5}$
$=\frac{1.5}{2.5}=0.6A$