question_answer

In the circuit shown the value of I m ampere is:

A)  1

B)  0.60

C)  0.4

D)  1.5

Correct Answer: C

Solution :

We can simplify the network as shown : So, net resistance, \[R=2.4+1.6=4.0\Omega \] Therefore, current from the battery, \[i=\frac{V}{R}=\frac{4}{4}=1A\] Now from the circuit , \[4I=6I\] \[\Rightarrow \] \[I=\frac{3}{2}I\] but\[i=I=I+\frac{3}{2}I=\frac{5}{2}I\] \[\therefore \] \[1=\frac{5}{2}I\] \[\Rightarrow \] \[I=\frac{2}{5}=0.4A\]