Electrical – Simple KVL Problem Not So Simple

LED's aren't best modeled as a pure resistor. As noted in some other answers, real LED's do have resistance, but often that's not the primary concern when modeling a diode. An LED's current/voltage relationship graph:

Now this behavior is quite difficult to calculate by hand (especially for complicated circuits), but there is a good "approximation" which splits the diode into 3 discrete modes of operation:

• If the voltage across the diode is greater than Vd, the diode behaves like a constant voltage drop (i.e. it will allow whatever current through to maintain V = Vd).

• If the voltage is less than Vd but greater than the breakdown voltage Vbr, the diode doesn't conduct.

• If the reverse bias voltage is above the breakdown voltage Vbr, the diode again becomes conducting, and will allow whatever current through to maintain V = Vbr.

So let's suppose we have some circuit:

^{simulate this circuit – Schematic created using CircuitLab}

First, we're going to assume that VS > Vd. That means the voltage across R is VR = VS - Vd.

Using Ohm's law, we can tell that the current flowing through R (and thus D) is:

\begin{equation} I = \frac{V_R}{R} \end{equation}

Let's plug some numbers in. Say VS = 5V, R=2.2k, Vd=2V (a typical red LED).

\begin{equation} V_R = 5V - 2V = 3V\\ I = \frac{3V}{2.2k\Omega} = 1.36 mA \end{equation}

Ok, what if VS = 1V, R = 2.2k, and Vd = 2V?

This time, VS < Vd, and the diode doesn't conduct. There's no current flowing through R, so VR = 0V. That means VD = VS = 1V (here, VD is the actual voltage across D, where-as Vd is the saturation voltage drop of the diode).

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.