Examples: Kirchoff's Current Law

Taking the current flowing into the junction as positive and that flowing away as negative we can obtain the following relationship:

0.2 + I_X = 0.3 + 0.15 A

(in other words, the sum of the current flowing into the junction is equal to the sum of the current flowing away from the junction)

Thus,

I_X = 0.3 + 0.15 - 0.2 = 0.25 A

First we can find the current flowing by applying Kirchhoff's Current Law at point A:

I_AB = 0.5 - 0.3 = 0.2 A

Next we can find I_X by applying Kirchhoff's Current Law at point B:

I_AB = I_X - 0.1 = 0.2 - 0.1 = 0.1 A

At first this might look difficult but it's just another Kirchhoff's Current Law problem! The maximum input current from the power supply is 6 A. We need to find the current taken by the remainder of the system, anything left over is available for powering any accessories that we might wish to use.

Total current demand is thus:

1.5 + 1.5 + 0.8 + 1.6 = 5.4 A

Available for the accessory socket is thus:

6 - 5.4 = 0.6 A (or 600 mA)