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 Solved Paper-2003
A)1 A
B)\[1.5\] A
C) 2 A
D)1 A
Correct Answer:
B
Solution :
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, \[V=lR\] \[\Rightarrow \] \[l=\frac{3}{2}=1.5\,A\]