question_answer

The current in the following circuit is :

A) 1 A                                         

B) \[\frac{2}{3}A\]

C) \[\frac{2}{9}A\]                                

D)  \[\frac{1}{8}A\]

Correct Answer: A

Solution :

                Key Idea: The currentin the circuit can be found by using Ohms law. From the given figure, the resistances in the arms AC and BC are in series. \[\therefore \]  \[R={{R}_{AC}}+{{R}_{BC}}=3\Omega +3\Omega =6\Omega \] Now R is in parallel with the resistance in arm AB, so           \[R=\frac{R\times {{R}_{AB}}}{R+{{R}_{AB}}}=\frac{6\Omega \times 3\Omega }{6\Omega +3\Omega }=2\Omega \] Therefore, the Ohms law can be stated as :                                 \[V=iR\] or                i = current \[=\frac{V}{R}\] Substituting the values, we get \[i=\frac{2V}{2\Omega }=1A\]                   (\[\because \]\[V=2V\]) NOTE: 1. If the current in the resistance flows in the direction ( as in case of charging of a battery), then          \[V=E-ir\] or            \[V<E\]                 2.If the current in the resistance flows in opposite direction, then \[V=E-ir\]                 or            \[V>E\]