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,
$R=24+1.6=4.0\Omega$
Therefore, current from the battery,
$i=\frac{V}{R}=\frac{4}{4}=1A$
Now from the circuit (b),
$4I^{\prime }=6I$
$\Rightarrow I^{\prime }=\frac{3}{2}I$
but $i=I+I^{\prime }=I+\frac{3}{2}I=\frac{5}{2}I$
$\therefore 1=\frac{5}{2}I$
$\Rightarrow I=\frac{2}{5}=0.4A$