# Confused about current flow

I am starting to learn about electronics and I am still confused about current flow. Initially the book would try to be clear about the distinction between conventional and actual flow but then further in the book the circuits are confusing me. Take e.g. the following DC circuit: Actual flow is of electrons pushing the other electrons from the − terminal to the + terminal. But then why is there 0 V between the − terminal and the first lightbulb? There has not been any resistance yet, so I would expect this to be 9 V. And then I would expect 0 V being at the + terminal.

Is my confusion clear? Could anyone then attempt to explain why my expectation is wrong?

• Forget about electron flow. Conventional flow is so called because that's what we normally use. Your diagram shows that the negative terminal of the battery has been selected as the reference 0V. Any conductor connected to it will be at 0V, so the left of the lower lamp is at 0V. Ohm's law tells us that when a current flows through a resistor (a lamp in this case) there will be a voltage drop - in this case the drop is 3V so the right hand connection to the lamp is at +3V. And so it continues at each lamp. – Chu Jun 10 '17 at 16:00
• Voltage is a concept which only makes sense when considering two different nodes. The voltage is the voltage DIFFERENCE between the two nodes. Most circuits have a ground which is, by definition, zero volts. All other voltages are understood, by convention to be measured with respect to ground. Forget about electron flow. Just use conventional current flow. It moves from battery positive terminal to battery minus terminal. You may want to remember that electrons have negative charge. – mkeith Jun 10 '17 at 17:21

## 1 Answer

IF each bulb is identical then the 9V is split evenly with 3V across each of 3 bulbs.

Also remember that Ground is an optional symbol to represent 0V ( not necessarily earth ground) at any point for voltage references and current flows from V+ towards lower voltages. Other nodes may be 0V too if there is 0 resistance or 0 current between them since the product is ($\Delta V=IR$) which means the voltage drop. But that is not relevant here as we expect current to flow in a loop shown in your schematic.

For KVL and KCL simple rules of calculation and also intuitive flow of current from a battery or any Voltage source, we use the convention for +current to flow from the +ve source and towards the -ve voltage.

Remember this is a LOGIC DIAGRAM or schematic and not a molecular physics diagram, so we EE's use this logic to speed up analysis with intuition. ( as some famous person once said, "it's all relative")

We don't think about how actual electrons collide with each other like a lineup Queue where one person bumps into the next and travels at the speed of light for a given dielectric surrounding the conductor with electrons (-) attracted towards the positive (+) conductor terminal between resistor with a shared current flow I=dQ/dt of charges.

... unless we want to examine molecular physics of static charges along with an AC or DC excitation that contribute to voltage breakdown before the actual breakdown arc (negative resistance) occurs. I only do this for failure analysis on double electric charge capacitors (Supercaps) , same as batteries and UHV transformer insulation failures (e.g.>20kV) but not for electronics per se.