# Current calculation

i am first year engineer and this is my first class on electronics. Can please someone explain why ix' = 0.2A ?

** I mean it is 0.2 if we add the resistors, but wouldn't it be 0.5 after the 6Ω?

• In situations like this it's easy to use Kirchoff's laws to determine what is happening. When doing so, you would see that at the junction between the 6 ohm resistor and the 9 ohm resistor has to have the same current, because the other branch doesn't actually go anywhere. If it was connected to ground, it would be a different situation; but in this case you really just have a single loop for the current to pass through.
– JMac
Jun 2 '17 at 12:10

In that circuit the resistors are in series, so the current through them is the same. In addition, to calculate the current, you can sum the values of the resistors (because they are in series) and then apply Ohm's law: $I=\frac{V}{R}=\frac{3V}{(9+6)\Omega}=0.2A$