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:

circuit schematic

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?

  • \$\begingroup\$ Knowing something about your background would help us give you a clear answer. For example, if you're a high school student who's currently learning algebra, we might give a different answer than if you're a grad student in physics. \$\endgroup\$
    – The Photon
    Aug 7 '13 at 23:21
  • \$\begingroup\$ I'm a high school student who's in a technical course for Informatics (That's how it's called here), so i learn programming and digital electronics, among other subjects. \$\endgroup\$ Aug 7 '13 at 23:24
  • \$\begingroup\$ Judging from all your other comments you are lacking understanding of some really basic concepts, which obviously makes it hard for you to draw conclusions. I suggest you read up on voltage and current sources, components in series and parallel, and the details about what voltage and current are in general. You really seem to have to start from ZERO. \$\endgroup\$
    – Rev
    Aug 8 '13 at 6:29
  • \$\begingroup\$ Is there any book or online text that can help me without introducing misconceptions? \$\endgroup\$ Aug 8 '13 at 14:53

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.


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.

  • \$\begingroup\$ I still don't get it when you say it HAS to be xVolts. Why if i connect the LED then, the voltage is around 3V? It gets higher with more load? That doesn't make any sense to me \$\endgroup\$ Aug 7 '13 at 23:33
  • \$\begingroup\$ Where are you measuring the voltage? You aren't actually increasing the load by adding more components in series, you're decreasing it. An increase in resistance is a decrease in current. Adding a diode (LED's can have up to 2 V drops, though 0.7 to 1.2 is more common) increases the apparent resistance. \$\endgroup\$
    – SLaG
    Aug 7 '13 at 23:41
  • \$\begingroup\$ Exactly, that's what i don't understand. If there's no resistor in the circuit, the current would be huge, but the voltage would... Be 10V on all points between +V and GND, right? If i add the resistor, i expected the current to be small, but the voltage to remain at 10V at all points. That's what i don't get. \$\endgroup\$ Aug 8 '13 at 0:13
  • \$\begingroup\$ If i measure the resistance from Arduino's +5V to it's GND pin, i get what i expected, 4.89V. But if i measure the voltage after the resistor to the same GND pin, i get 0V. \$\endgroup\$ Aug 8 '13 at 0:15
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    \$\begingroup\$ The voltage doesn't remain the same throughout the circuit; it drops after the components that drop it. The current remains the same throughout the circuit. \$\endgroup\$ Aug 8 '13 at 0:25

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