Hi All,
Please excuse this simple question, I'm new to electrics/electronics and teaching myself...........
A question about voltage drop.......

When we talk about a voltage drop across something, do we mean the voltage has dropped TO - or voltage has dropped BY.
For example, in a series circuit I have a 9v battery and when I put in a resistor of q ohms and measure the volts it reads 1.5v.
Does that mean there is 7.5v through the resistor or 1.5v?

I think it means the voltage has dropped BY because the series rule is: "total voltage in a series circuit is equal to the sum of the individual voltage drops". But please confirm,
Thanks,
Steve

 

A drop across is measured... ACROSS the component.

A voltage dropped to (probably) means the voltage from that point to a reference point (such as "ground").

If you just have a 9V battery and a resistor then unless you are seriously overloading the battery then the 9V of the battery is also across the resistor... you need another part to get another drop.

 

Ohm's Law is usually in charge of voltage drop, but there are exceptions. Silicon diodes, for example, drop 0.6 to 0.7VDC. LEDs are extremely variable, but pretty constant for an individual component. Zeners, though their mechanism is different, are much the same.

 

Here's a pic and some math to help you. First, calculate the current. 9V/ 300 ohms = .03A This is an application of Ohms law E=IR Here, it is I=E/R

.03A x 200 ohms = 6 volts
.03A x 100 ohms = 3 volts
3 volts + 6 volts = 9 volts (your source.)

 

Hi Again,
Thanks for all your answers.
Perhaps I'm being too pedantic when I see the phrase "voltage drop"...
I've attached a short Word doc to explain what I'm trying to say.....
Had to do it like this because I couldn't upload the images, so if you could have a quick look, I'd be grateful. Hope this makes it clearer what I'm trying to ask.
Or am I looking at it from the wrong angle?
Thanks,
Steve

 

As I understand it, your first resistor drops 1.5V so then 7.5V exits to the second resistor, and drops 5V, 2.5V exits and is dropped by the third, to 0V.

do the resistors "used up" the voltage.

Notice MBo's post where the current through the resistors was the same for each resistor. So your current times the voltage drop will give you the watts

 

You need to internalize two key things. First, voltage is always measured with respect to something else. Saying that point A has a voltage of 7 is meaningless unless it's referenced to something else. Voltage is always expressed as a potential difference between two points. Think of measuring a voltage in a circuit -- you'll always measure with two leads from a voltmeter. If you measure the voltage with the leads on either side of a resistor, that's the voltage drop across the resistor and is by virtue of the current flowing through the resistor.

Now, if you put one voltmeter lead at another point in the circuit and measure on either side of the resistor with the other lead, you'll get two numbers. Those are the voltage difference between the respective sides of the resistor and the other point in the circuit. If you subtract those two numbers, you'll get the voltage drop across the resistor. This is an easy experiment to do, so I suggest you do it.

The second key idea is that if you sum the voltage drops around a closed loop in a circuit, the sum is zero (and this assumes no changing magnetic flux through the circuit). This is usually called Kirchoff's voltage law, but it's really only a statement of the conservation of energy. Kirchoff's current law expresses the notion that the sum of the currents into any node in a circuit is zero, but it is just a statement of the conservation of charge.

 

And the verbage I use is "the voltage drop is", and a condition of that part.