# In the circuit shown, the current passing through the:

$\begin {array} {1 1} (A)\;3 \: \Omega \: resistor\: is\: 0.50\: A & \quad (B)\;3 \: \Omega \: resistor\: is\: 0.25\: A \\ (C)\;4 \: \Omega \: resistor\: is\: 0.50\: A & \quad (D)\;4 \: \Omega \: resistor\: is\: 0.20\: A \end {array}$

So the current through the 9V battery is 1A.
Also note, that the current through the 8 $\Omega$ branch nearer to the battery,
is 0.5A and through the second 8 $\Omega$ branch is $0.5 \times 0.5 = 0.25A.$
Hence, the current through the 4$\Omega$ resistor is 0.25A.
Ans : (D)
edited Mar 14, 2014