question_answer

Two resistors of resistances\[2\Omega \]. and \[6\Omega \]. Are connected in parallel. This combination is then connected to a battery of emf 2 V and internal resistance\[0.5\Omega \]. What is the current flowing through the battery?

A)  \[\frac{4}{17}A\]

B)  \[4A\] 

C)  \[1A\]

D)  \[\frac{4}{3}A\]

Correct Answer: C

Solution :

Resistors \[2\Omega \], and \[6\Omega \], are connected in parallel. Their equivalent resistance is \[\frac{1}{{{R}_{P}}}=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\] \[{{R}_{P}}=\frac{2}{3}\Omega =1.5\Omega \] Current flowing through the battery is \[I=\frac{\varepsilon }{{{R}_{P}}+r}=\frac{2V}{1.5\Omega +0.5\Omega }=1A\]