Why does the voltage changes in this simple circuit?

Question

I'm beginning to learn electronics, and there's something I just can't grasp. I was told that a high resistance in a circuit would lead to a smaller current passing through it, but the voltage would remain the same. Now I'm facing a puzzle as what I learned and what I see in the real world are very different from each other. As example:

Upon seeing the circuit, I thought the current would flow from +10V to GND, at 0.01A. Actually, as you see in the image, the voltage kinda "stops" in the resistor. Trying this on an arduino, and if I put a LED in series with the resistor, I get about 3V. I really don't get what's happening here, can someone give me a light?

Answer 1

Remember that, in circuit analysis, "wires" are simply a tool we use to show connected components. You should try to see circuits as a combination of nets; a net simply describes how component pins are connected to each other. I know there can only be two different voltage readings in your circuit because your circuit only has two nets: ground, and 10V. Think about it. Every pin of every component in this circuit (in this case, a voltage source and resistor) is hooked up to only those two nets. No other nets exist.

Remember, voltage is always measured between nets, so there's really only one voltage to measure in your circuit, since you only have two nets -- ground, and 10V. It makes no sense to talk about the voltage at different points along a wire; just imagine the wire didn't exist, and the component(s) were hooked directly to each other. For example, if you hooked the little ground symbol straight up to your resistor (without the wire labeled "B"), that wouldn't affect the operation of the circuit at all.

Explanation: voltage doesn't change in an ideal wire. In other words, if you were to build that circuit, and measure the voltage (relative to ground) at every point along the circuit, you'd measure 10V all along the wire that hooked your positive voltage source to the resistor. Then, on the "bottom side" of the resistor, you'd measure 0V, all along the wire that hooked the resistor to ground.

Voltage only gets dropped across loads. Resistors, incandescent bulbs, motors, LEDs (and other semiconductor devices), etc.

So, if you added a second 1k resistor in series below the one you already have, you'd (again) measure 10V at the top of the first resistor, then you'd measure 5V in between the two resistors, and then you'd measure 0V at the bottom of the second resistor.

NB: We're talking about ideal wires here. In reality, every wire is a combination inductor/capacitor/resistor. It drops voltage. It stops sudden changes in current. It stops sudden changes in voltage. Wires are nasty things!

Also, you may have gotten confused by the "edge cases" that violate circuit theory; For example, if you replaced the resistor in your circuit with a wire, it would appear that the wire would have to drop 10V. But that circuit is invalid, and you can't apply analysis techniques to it.

Answer 2

What you're seeing is called voltage drop. The circuit is acting as you thought it should act, though you're current is out.

Because you are using a voltage source, the 10V side of the resistor HAS to be 10V because you are applying the 10V. The 0V side of the resistor HAS to be 0V as this is the reference point for the circuit.

Because you have defined the voltage (10V) and the resistance (10kohm), the dependent variable is current as defined by ohms law (V=IR).

Re-arranged, I (current) = V/R Substitute, I = 10 / 10000 Answer I = 0.001 A or 1 mA

Now if you were to have 2 x 10 kohm resistor in series and then apply 10 V, the point between the 2 resistors would be 5 V but the 10 V side would be 10 V and the 0 V side would be 0 V because that's how they were defined.

All the voltage has to be "used up" when you are using a voltage source. It's like jumping off a brick wall that is say 1m above the ground. The top of the wall is 10 V, the ground is 0V, the difference in height between the 2 is what gives you the ability to jump (potential difference). The action of you going from high to low is current flow.

Regarding the arduino, it depends where you measure the voltage. If you are using 2 pins on the board itself, it may have a resistor connected between the pin and ground that protects it. That's why you would see 3V, like the 2 resistors above.