question_answer

4 cells each of emf 2V 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)  2 A                                        

B)  1.5 A

C)  1A                                         

D)  0.888 A

E)  0.75 A

Correct Answer: D

Solution :

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