Question: In a linear voltage regulator like a TO-220: does Kirchhoff current law apply?

What is the equation for the three currents flowing?

In other words, what is the relation between I1, I2, I3 in the below circuit?
(by I2 I mean the current flowing from pin 1 to pin 2 (GND) )

And is I2 a fixed value? How can I find I2 in data sheet?

[![enter image description here](https://i.stack.imgur.com/Mia4y.png)](https://i.stack.imgur.com/Mia4y.png)

Answer: 
I'm going to make a few assumptions here, I'm going to assume that by I2 you mean the current flowing to GND from the voltage regulator.

In a linear voltage regulator (I've used an L7805 as the example for this answer) your output current is always going to be:

![I3 = I1 - I2](https://latex.codecogs.com/gif.latex?I_%7B3%7D&space;=&space;I_%7B1%7D&space;-&space;I_%7B2%7D)

Where I3 is your output current, I1 is your input current and I2 is the current flowing to ground, also known in the datasheet as the **quiescent current**. This quiescent current value changes slightly depending on your application but it is mainly a set value. For the case of an L7805 it is ~6mA.

Quiescent current is defined as the difference between the input an output currents. This is the value that I think you're after. Lower the quiescent current, higher the efficiency, although apart from specific or close to tolerance applications it is usually considered negligible
As the 78xx series is based on a linear regulator design, the input current required is always the same as the output current. As the input voltage must always be higher than the output voltage, this means that the total power (voltage multiplied by current) going into the 78xx will be more than the output power provided. The difference is dissipated as heat. This means both that for some applications an adequate heatsink must be provided, and also that a (often substantial) portion of the input power is wasted during the process, rendering them less efficient than some other types of power supplies. When the input voltage is significantly higher than the regulated output voltage (for example, powering a 7805 using a 24 volt power source), this inefficiency can be a significant issue.