# Why am I always calculating Voltage Drop, instead of "remaining voltage"

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#### rokuez

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I am plowing through this teach yourself electronics book & anytime I have to calculate anything to do with voltage it is always voltage drop.

Why don't I ever have to calculate how much voltage is left after the drop?

I asked a friend this and he said there "is no voltage left over". He gave an example of a basic series circuit with a 10 V power source, and three 1k Ohm resistors connected in series. I don't quite get what he means by no voltage left over...? Does he mean specifically that no electrons are entering the positive terminal of the power source i.e that they are all used up by the components?

I'm doing simple calculations with series circuits, but I'm just really curious. Is voltage drop all the matters i.e voltage reduction how much voltage was lost? What about the other voltage / potential difference after the drop across each component? Say we have 10 volts, and a drop of 1 volt. why am I always looking for that 1 volt drop, and not the remaining 9 volts left over so to speak.

I think of voltage as pressure so maybe I'm not conceptually understanding voltage. I think why not know the amount of electron/charge pressure so to speak after the pressure-reduction, or voltage drop.

Second part of my question kinda off on a little bit of a tangent...
Is it possible to measure voltage at one specific point. Or must voltage drop always be measured across a component?

Three 1k resistors "in series" on a 10v supply.
The reason why we are interested in "voltage drop" is this: The voltage drop (across the resistor in question) will cause heating and we need to know what wattage is needed for the resistor.
When you work out the voltage across the first resistor, then the second resistor and finally the third, the result will be the addition of three values that equal 10v. This is what he means by "there will be no voltage left over."

Suppose you want to measure the voltage at a particular point in a circuit.
You put the positive probe on the point.
Where do you put the negative probe?
You can put the two probes anywhere on a circuit. The result will be a voltage across maybe one component, two components or more.

Voltage is shared between all components that resist electron flow. The battery pushes one way, and everything else pushes the other way. If you had some left over voltage, it would boost the batteries voltage more and create a voltage gain, which would be odd This was already said, but I'm going to say it again. The voltage drop of all components on an electrical path will equal the supply voltage. All voltage is dropped by the time you reach the negative terminal. This is what your friend was referring to.

Electricity flows through the path of a circuit all at once. It does not have a leading or trailing edge to it. You can only somewhat compare it to a closed-loop (man, that word is coming up a lot lately) water system, not an open loop. As soon as you start pumping at one end of the system, the other end must start moving too.

Voltage drop is a basic unit used in Ohm's law.

It allows you to determine the other elements of current in amperes, resistance in ohms and power in watts.

Second part of my question kinda off on a little bit of a tangent...
Is it possible to measure voltage at one specific point. Or must voltage drop always be measured across a component?

The voltage at any point must be measured to a reference, most often "ground" or "common." Voltage measured across a component is the so-called "voltage drop" across that component.

I am plowing through this teach yourself electronics book & anytime I have to calculate anything to do with voltage it is always voltage drop.

Why don't I ever have to calculate how much voltage is left after the drop?...
Oh but you do. Suppose you have a simple circuit consisting of a battery in series with a resistor and a lamp. You have a 9V battery, but a 6V lamp. You don't want to burn out the lamp, so you use a resistor to "drop" 3V. What is left is the 6V needed for the lamp.

But it is just as valid to ask "what is the drop across the lamp" Think of it this way: in a series loop, the power source (battery) is a "rise". All power-consuming "loads" are a "drop". Paraphrasing Kirchhoff's voltage law (KVL); the sum of "rises" and "drops" in a circuit loop is ZERO. If you put a + in front of "rises", and - in front of "drops", in our example +9 -3 -6 = 0, or 9 = 3 + 6. Last edited:
Why don't I ever have to calculate how much voltage is left after the drop?

Maybe you haven't yet, but you will at some point. You'll need to be able to use that to figure out other voltages in the circuit. The way we were taught was around Kirchoff's Voltage Law, which basically said that the sum of voltages in any "loop" including the source would be 0. I've been working on analyzing a BJT transistor, and knowing the voltage 'left over' is helpful (at least in my circuit) for figuring out the Base-Emitter voltage and the Collector-Emitter voltage that are "left over" after resistors' voltage drop.

So, if you have more than one element in the circuit in series, the voltage "left over" for the 2nd (or 3rd or 4th) element in series can be useful to know.

In your friend's example, with 3 identical resistors in series, they will each drop 1/3 of the total source voltage. Because they drop, in total, the full 10V across all of them. So Resistor 1 has 10V "left over" since it is hooked up to the batt. It has a 3.33V voltage drop. Resistor 2 has 10V-3.33V - 6.66V "left over". It also drops 3.33V. So Resistor 3 has 6.66V - 3.33V = 3.33V "left over". It drops the remaining 3.33V so that there is 0V "left over".

The energy potential from the battery is always all used up by the stuff connected across the + terminal to the - terminal of the battery. The fact that you connect stuff to the - terminal means that there will be zero energy at that point.

You can only measure voltage across two points... one of those points might be ground, which is sort of the 'voltage left over' concept. Or you can measure across two points to measure the 'drop'

Michael

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