question_answer

In the circuit shown the value of \[I\] in 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,                 \[R=2.4+1.6=4.00\] Therefore, current from the battery,                 \[i=\frac{V}{R}=\frac{4}{4}=1\,A\] Now from the circuit (b),                 \[4I=6I\] \[\Rightarrow \]               \[I=\frac{3}{2}I\] but \[i=I+I=I+\frac{3}{2}I=\frac{5}{2}I\] \[\therefore \]  \[1=\frac{5}{2}I\] \[\Rightarrow \]               \[I=\frac{2}{5}=0.4\,A\]