question_answer

A 3 V battery with negligible internal resistance is connected in a circuit as shown in the figure. The current I, in the circuit will be      [AIEEE 2003]

A)             1 A

B) \[1.5\] A  

C)             2 A

D) 1 A

Correct Answer: B

Solution :

[b] Resistance in the arms AC and 6C are in series,

\[\therefore \]                  \[R'=3+3=6\,\Omega \]

Now, R' and \[3\,\Omega \] are in parallel.

\[\therefore \]      \[{{R}_{eq}}=\frac{6\times 3}{6+3}=2\,\,\Omega \]

Now,     \[V=lR\]

\[\Rightarrow \]   \[l=\frac{3}{2}=1.5\,A\]