Question

In the circuit shown the value of I in ampere is

• A. 1
• B. 0.60
• C. 0.4
• D. 1.5

Answer 1

Answer: (C)

We can simplify the network as shown
So, net resistance, $R=2.4+1.6=4.3$ 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 $i=I+I=I+\frac{3}{2}I=\frac{5}{2}I$ $\therefore$ $1=\frac{5}{2}I$ $\Rightarrow$ $I=\frac{2}{5}=0.4A$