# Voltage drop through a resistor using Ohm's Law?

I'm having a very hard time comprehending basic Ohm's Law.

I was recently watching a video on the operation and use of Operational Amplifiers, and was completely lost because I couldn't understand what the resistors in the circuit were actually doing.

For example, part of his circuit was limiting 10V to 5V using a 2.2k resistor.

Where do the values go in Ohm's equation? How can we know the output voltage without the current in amperes?

The only two configurations I can think of:

$$10V = \frac{2.2k\Omega}{I}$$ $$5V = \frac{2.2k\Omega}{I}$$

But, he got the value without knowing the amperage, how is this accomplished?

• If that (dropping 5V) was realy the purpose of that resistor it could do that only at a fixed current (which you can calculate). – Wouter van Ooijen May 12 '15 at 6:22
• Maybe, you're speaking about a voltage divider. Connect two resistors of same size in series to 10V, and you get 5V between them. (measured from Ground/0V) – sweber May 12 '15 at 6:44
• If you could please copy the full circuit, I doubt it is only a resistor, it must be two (a voltage divider). – davidrojas May 12 '15 at 6:44
• "he got the value" - what value AND please don't ask me to watch the video and pause somewhere. – Andy aka May 12 '15 at 7:34

## 2 Answers

I was mistaken. He was using a voltage divider.

This is the equivalent piece of the circuit I was concerned about... simulate this circuit – Schematic created using CircuitLab

Because I didn't understand exactly how current flows, in my mind ONLY THE FIRST RESISTOR had any effect on the voltage of the output wire because logically current had only gotten to that point. The following terms may not be correct, but they suffice to explain the concept.

What I realize now is that the TOTAL RESISTANCE of everything leading back to ground dictates voltage as well. CURRENT RESISTANCE dictates the amount of CURRENT at the point of measurement.

The following is how you would get the output voltage, if it were unknown:

I = Amperage
V = Voltage
R = Resistance

Ohm's Law: $I = V/R$

Total Resistance (TR) (The Total Resistance in Your Circuit) = R1 + R2
Current Resistance (CR) (The Resistance in Your Circuit Up To The Point of Measurement) = R1

TR = 4.4k
CR = 2.2k

This will be the current in mA that flows through our circuit:

$I=V/R$

$I = 10v/TR$
$I ≈ 2.27mA$

This will be the voltage drop across the resistance up to our measurement point. (I.E. Our Output):

It is completely coincidental that our voltage drop is our desired voltage. At this stage we have out voltage drop not the actual output voltage.

$V=IR$

$V = .00227A * CR$
$Voltage Drop (VD) ≈ 4.994v ≈ 5v$

This will be the voltage on our measurement point. (I.E. Our Output):

$Output Voltage (OV) = V - VD$
$OV = 10v - 5v$

$OV = 5v$

Ohm's Law only talks about passive components: resistors, capacitors, and inductors. When you put an active device into the mix, Ohm's Law is insufficient.

That's not to say that Ohm's Law is useless when working out what an op-amp circuit will do, but that it does not by itself tell you what the circuit does. This is why there are standard equations for op-amp behavior, such as the noninverting and inverting gain equations.

When you're dealing with an op-amp circuit other than simple gain blocks, you either need to do your own linear circuit analysis or you need to find an analysis of the same sort of circuit. Linear circuit analysis books are big and heavy for a reason, and there are multiple good books on op-amps for the same sort of reason.