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 \[V>E\]You need to login to perform this action.
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