Question

4 cells each of emf 2 V and internal resistance of $1\Omega$ are connected in parallel to a load resistor of $2\Omega$ .
Then the current through the load resistor is

• A. 2A
• B. 1.5A
• C. 1A
• D. 0.888A
• E. 0.75A

Answer 1

Answer: (D)

The combined resistance of combination is
$R^{\prime }=R+\frac{r}{4}=2+\frac{1}{4}=\frac{9}{4}$
$V=2volt$
$\therefore V=iR$
$\Rightarrow i=\frac{V}{R}=\frac{2\times 4}{9}=\frac{8}{9}$
$=0.888A$