# How does one determine the voltage of a node?

This sounds like a very straightforward question, and probably has an obvious answer. I've read a section in a physics book on electricity, I've read a textbook on DC and this simple concept seems to elude me.

Say I have this circuit (The one to the left): simulate this circuit – Schematic created using CircuitLab

It's about as simple as it can get. A voltage source connected to a resistor.

I've chosen these values for simplicity's sake. It's easy enough to use Ohm's Law to get the current running through the circuit:

$$I= \frac{V}{R}$$
$$I= \frac{1V}{100Ω}$$
$$I= .01A = 10mA$$

Okay, so now we know at any point we're getting 10mC/s through any point in the circuit.

But, with that interpretation, there is no voltage loss anywhere. If there is the same current everywhere, and the same circuit resistance everywhere, there is the same voltage...everywhere. Regardless of whether you're below, above, or parallel to whatever component your asking about.

But if this is true, voltage dividers wouldn't work, and current shunts wouldn't drop 100% of the voltage.

Ohm's Law seems counter-intuitive.

This circuit is a current shunt. It drops 100% of the voltage across it. That means that Node2 should be at, or electrically common to ground.

But, let's take the second circuit into consideration. Suppose SW1 is on. D1 will be on for short period of time until the capacitor is fully charged. When I disconnect the switch D1 will turn on through the charge stored on C1 at the rail voltage.

How is this possible? I'm having a very hard time understanding this simple concept...

• 1V in your text, 10V in the diagram. Not a big deal, but you might want to correct. – BowlOfRed Jun 7 '15 at 6:14
• What do you mean by "and the same circuit resistance everywhere"? You have 0 ohm from voltage source to node1, 100 ohm from node1 to node2, and 0 ohm from node2 to voltage source. – BowlOfRed Jun 7 '15 at 6:17

One very important thing to keep in mind is that voltages are always referenced between two points. Ohm's law states the voltage across a resistor is proportional to the current flowing through it.

So in case 1 (left circuit), let's assume the current flows from node1 to node2. The voltage across the resistor then is $V_{node1} - V_{node2}$. Because we have the ideal voltage source V1, this voltage difference is by definition 10V. The current flowing through it can be computed to be 10mA (as you have done).

Very clearly, the voltage at node1 and the voltage at node2 are not equal. We just showed that the voltage at node1 must be 10V higher than the voltage at node2.

Additionally it is just as wrong to say that the voltage at node 1 is 10V. It could be 100V if $V_{node2} = 90V$, or it could be -5V if $V_{node2} = -15V$. The only time it makes sense to say what the voltage at a node is if you define a reference node (also known as a ground node), which we define to be at 0V.

In case 1, there are two possible choices for the ground node: simulate this circuit – Schematic created using CircuitLab

These two circuits are identical to your left circuit, I've simply defined a my ground node differently between the two.

Now and only now does it make sense to say what the voltage at a node is.

\begin{gather} V_{node1} = 10V\\ V_{node2} = 0V\\ V_{node3} = 0V\\ V_{node4} = -10V \end{gather}

In case 2 (your right circuit), we still have that $V_{node1} - V_{node2} = I R_1$ (by Ohm's law), but because the voltage source is no longer connected to node1 and node2, $V_{node1} - V_{node2} \neq 10V$ (well, technically there is a time this is true, but that time is extremely short and dictated by other factors). Instead, when the switch is closed V1 provides the constraint that $V_{node1} - V_{node3} = 10V$ (assuming that node3 is the very bottom node). simulate this circuit

We need additional equations relating the voltage at node2 to the other nodes before we can determine the behavior of the circuit.

As alluded to by Anklon's comments, with SW1 closed the capacitor will charge up until the diode begins to conduct. This happens nominally at $V_{node2} - V_{node3} \approx 0.7V$. We now have enough equations to solve for the steady state behavior of the circuit:

\begin{gather} V_{node1} - V_{node2} = I R_1\\ V_{node2} - V_{node3} = 0.7V\\ V_{node1} - V_{node3} = 10V \end{gather}

You'll find that $V_{node1} - V_{node2} = 9.3V$.

( I don't see any relation between your title and your question.Try to give relevant title. )

We assume the forward bias voltage of $D_1$ is 0.7V . So when you keep $SW_1$ close current flow through the diode at 0.7V difference and So the capacitor is also charged having 0.7V difference. simulate this circuit – Schematic created using CircuitLab

When you open $SW_1$ capacitor is still charged and it's voltage is 0.7V which is forward bias voltage for diode.So the voltage drop across the Resistor will be (10-0.7) Volt . So current will go thorough the diode for a certain time. In time , capacitor will loose charges and when it's voltage came down below the forward bias voltage ,Current will stop flowing or negligible amount .

• Thanks for answering, but I wasn't really questioning the operation of the capacitor. I'm trying to figure out how you can have a "100%" voltage drop...yet still have a current. The capacitor was just to prove that I still get current at the rail voltage even after the "100%" voltage drop. – Allenph Jun 7 '15 at 5:05
• When $SW_1$ is on at steady state condition, voltage drop across the resistance will be (10-0.7)Volt. Your current will be (10-0.7)/100 = 93mA . So your total voltage drop 9.3 + 0.7 = 10 Volt .Capacitor just stored some charges and when $SW_1$ off, it release it .This charge flow creates this 0.7 voltage difference and it drops when passing through the diode. – Anklon Jun 7 '15 at 5:10