# Confused Circuit Math with Resistors and Voltage?

#### shintashi

Joined Apr 11, 2016
6
I think I forgot something important somewhere, and here's the basic dilemma of math:

Ohms law I = V/R, V = IR

So if I have 10 volts and want to run something at 5 volts, I cut my resistance in half?
because if i have 2 amps and 5 ohms, I have 10 volts? and 2 amps = 10 volts / 2 ohms

But then I noticed when doing the calculations, in a series circuit, if I have 3 LEDs or Motors,
I might start with 10 volts, then subtract chunks of voltage at each LED/Motor, so that the
second motor only has say, 7.5 volts left, and then the third motor might only have 5 volts left.

But with a resistor, how do I subtract 5 volts directly, so that my 10 volt power supply only gives 5 volts to my motor or LED?

If i know my input voltage, and my resistor's ohms, such as 330 ohms, how do I get to 5 volts on the other side?
With 10 volts, its saying a 330 resistor should have .03 amps, but that doesn't tell me how to change my voltage.
If i have a 330 resistor in place and its reading 10 volts, then if...

Now I'm confused. How do I get subtraction when its all multiplication and division? And why does it look like V = IR is saying by cutting my resistor from 330 to 165, I go from 10 volts to 5? Now I'm even more confused.

#### HermanB

Joined Apr 18, 2016
7
I have similar issue. Please can someone help?

#### MrChips

Joined Oct 2, 2009
29,271
Think overall load or total series resistance.

In order to drop the voltage to a load RLOAD we need a series resistance RSERIES.

RTOTAL = RSERIES + RLOAD

Now we can apply Ohm's Law.

I = V/R
TOTAL

The voltage across the load is V x R

The voltage across the series resistor is V x R
SERIES/RTOTAL

#### wayneh

Joined Sep 9, 2010
17,406
So if I have 10 volts and want to run something at 5 volts, I cut my resistance in half?
No, it means that your "something" will see 5V if you have two "somethings" in series. Using a resistor that has a similar resistance to the something will also accomplish that. So the total resistance has doubled, and the current through the two somethings in series supplied 10V will be the same as just one something supplied 5V.
...because if i have 2 amps and 5 ohms, I have 10 volts? and 2 amps = 10 volts / 2 ohms
That 2nd part is not right.

But then I noticed when doing the calculations, in a series circuit, if I have 3 LEDs or Motors,
I might start with 10 volts, then subtract chunks of voltage at each LED/Motor, so that the
second motor only has say, 7.5 volts left, and then the third motor might only have 5 volts left.

But with a resistor, how do I subtract 5 volts directly, so that my 10 volt power supply only gives 5 volts to my motor or LED?
You're close. Everything in series will have the identical current passing through it. Across all the devices, the voltage will drop 10V (your total). The amount of drop across each item will depend on its own properties. For resistors, the ∆V will simply be in proportion to the resistance value. Loads like motors and LEDs are more complicated because they do not have a constant resistance or voltage drop.
If i know my input voltage, and my resistor's ohms, such as 330 ohms, how do I get to 5 volts on the other side?
With 10 volts, its saying a 330 resistor should have .03 amps, but that doesn't tell me how to change my voltage.
If i have a 330 resistor in place and its reading 10 volts, then if...
Yeah, you're confusing things. Perhaps the most critical one is that a resistor drops voltage in proportion to current. It is not a constant voltage drop. Think of it as a partly-open valve on a water supply. To get more current through it, you have to supply more pressure (voltage), and you will see a greater pressure drop across it. If there is barely any current flowing through a resistor, it will not be dropping much voltage.

#### shintashi

Joined Apr 11, 2016
6
yeah, 2 amps = 10 volts / 2 ohms should have read 2 amps = 10 volts / 5 ohms, but i think i got lost somewhere while typing.

MrChips post looks correct but I didn't understand how to use the information to lower my voltage

Wayneh - is there a way to get resistors to lower voltage in the same way motors and LEDs can in series? I really don't want to burn out my stuff. I thought all resistors had some kind of rated amp use or something?

#### shteii01

Joined Feb 19, 2010
4,644
yeah, 2 amps = 10 volts / 2 ohms should have read 2 amps = 10 volts / 5 ohms, but i think i got lost somewhere while typing.

MrChips post looks correct but I didn't understand how to use the information to lower my voltage

Wayneh - is there a way to get resistors to lower voltage in the same way motors and LEDs can in series? I really don't want to burn out my stuff. I thought all resistors had some kind of rated amp use or something?
You can do it in a couple of ways:
1. Voltage Divider
2. Stick a couple of diodes in series, "normal" diodes use 0.7 volts, so each diode will use 0.7 volts. Lets say you need to go from 5 volts to 2.5 volts, stick three diodes in series (3*0.7 volts=2.1 volts), 5-2.1=2.9 volts, not perfect, but it is probably good enough/close enough to 2.5 volts without stressing stuff too much.

#### HermanB

Joined Apr 18, 2016
7
Think overall load or total series resistance.

In order to drop the voltage to a load RLOAD we need a series resistance RSERIES.

RTOTAL = RSERIES + RLOAD

Now we can apply Ohm's Law.

I = V/R
TOTAL

The voltage across the load is V x R

The voltage across the series resistor is V x R
SERIES/RTOTAL

#### HermanB

Joined Apr 18, 2016
7
Thanks. So far, I follow. My problem is more complex, I think. I will draw a simple schematic, and then post it with my problems/questions, and will appreciate any help to solve a real present and quite urgent issue.

#### shintashi

Joined Apr 11, 2016
6
so a voltage divider is one of the ways to lower voltage with resisters, rather than the 0.7 diode trick? I found this video:

the 27 second mark appears to do what I need it to for multiple items in a series, while the 11-20 second mark appears to use equal ohm resistors spaced apart to produce 50% voltage between the two. I have not actually seen what a voltage divider was until today

P.S. Is there any benefit to using higher or lower ohm resistors? Or is their maximum amp/overheating rating the only thing that matters?

#### wayneh

Joined Sep 9, 2010
17,406
You want to choose the value of the resistors to be as high as possible, to minimize current and wasted power, while ensuring that whatever is drawing power from the divider is drawing no more than ~10% of the current. So sometimes you need to use lower values in the divider to increase the current flowing in the divider, so that the load current remains a small distortion. At some arbitrary level, you need to add the load current into the calculations.

You really should share what you are trying to do. You'll get specifics instead of generalities. You may need a proper voltage regulator. They're wonderful, so don't avoid them.

#### hp1729

Joined Nov 23, 2015
2,304
I think I forgot something important somewhere, and here's the basic dilemma of math:

Ohms law I = V/R, V = IR

So if I have 10 volts and want to run something at 5 volts, I cut my resistance in half?
because if i have 2 amps and 5 ohms, I have 10 volts? and 2 amps = 10 volts / 2 ohms

But then I noticed when doing the calculations, in a series circuit, if I have 3 LEDs or Motors,
I might start with 10 volts, then subtract chunks of voltage at each LED/Motor, so that the
second motor only has say, 7.5 volts left, and then the third motor might only have 5 volts left.

But with a resistor, how do I subtract 5 volts directly, so that my 10 volt power supply only gives 5 volts to my motor or LED?

If i know my input voltage, and my resistor's ohms, such as 330 ohms, how do I get to 5 volts on the other side?
With 10 volts, its saying a 330 resistor should have .03 amps, but that doesn't tell me how to change my voltage.
If i have a 330 resistor in place and its reading 10 volts, then if...

Now I'm confused. How do I get subtraction when its all multiplication and division? And why does it look like V = IR is saying by cutting my resistor from 330 to 165, I go from 10 volts to 5? Now I'm even more confused.
Applying 10 V. You want the motor to have 5 V. You need to add a resistor in series with the same resistance as the motor to drop the same voltage. What is the typical current the motor will draw at 5 V? R = V/I.
Reality can get more complicated.

#### wayneh

Joined Sep 9, 2010
17,406
I'll add a rule of thumb that I use: Resistor dividers are for information signals, regulators are for power signals. The distinction between information and power is not a hard line, but if voltage is what matters and current is just a nuisance, it's information. If current is important and more than used by, say, an LED, then it's power and needs something more than a resistor divider.

#### BobaMosfet

Joined Jul 1, 2009
2,082
I think I forgot something important somewhere, and here's the basic dilemma of math:

Ohms law I = V/R, V = IR

So if I have 10 volts and want to run something at 5 volts, I cut my resistance in half?
because if i have 2 amps and 5 ohms, I have 10 volts? and 2 amps = 10 volts / 2 ohms

But then I noticed when doing the calculations, in a series circuit, if I have 3 LEDs or Motors,
I might start with 10 volts, then subtract chunks of voltage at each LED/Motor, so that the
second motor only has say, 7.5 volts left, and then the third motor might only have 5 volts left.

But with a resistor, how do I subtract 5 volts directly, so that my 10 volt power supply only gives 5 volts to my motor or LED?

If i know my input voltage, and my resistor's ohms, such as 330 ohms, how do I get to 5 volts on the other side?
With 10 volts, its saying a 330 resistor should have .03 amps, but that doesn't tell me how to change my voltage.
If i have a 330 resistor in place and its reading 10 volts, then if...

Now I'm confused. How do I get subtraction when its all multiplication and division? And why does it look like V = IR is saying by cutting my resistor from 330 to 165, I go from 10 volts to 5? Now I'm even more confused.
It's OK. You're confused because how voltage, resistance and current work together simultaneously is still confusing to you. This isn't your fault, it's actually a failing of how electronics is taught, documented, and written about. The entire education system has a *lot* to learn about how this stuff actually should be presented.

• Why all division or mutliplication? Because Ohm's law describes a relationship (or RATIO) between current and resistance to the flow of that current. None of the 3 values happen in singular, they all operate together. Ohm's law is a 'pie' or 'circle' formula. E = I * R, is the same as I = E/R, is the same as R = E/I. It's that last one you're interested in.
• Confusing, sure. Everything must be accounted for-- but when you're new to electronics, it isn't obvious what things actually mean. Even Ohm's law is not actually understood by most people in electronics. Voltage isn't the amount of voltage present (when you're talking about anything other than a power source), it's the amount of voltage necessary (dropped) to be used in order to cause an interaction. Understanding this fundamental piece will help you more than almost any other.

So, using your example, you're saying I have a 10 Volt power-supply and I want to drop (or lose) 5 volts across a load and I don't want the load drawing more than 2 Amps max. That becomes:

R = (10-5)/2
R = 5/2
R = 2.5 Ohms.

Note the first line-- The '(10-5)' part is where we are saying we are splitting the max available voltage between the resistor, and the load. Each gets 5 volts.

Now, you can't find a 2.5 Ohm resistor very easily, so likely you'll have to use a 3.3 Ohm resistor, which means you'll wind up with something less than 2Amps allowed, max for the load. But for sake of simplicity, we'll assume we have a 2.5-Ohm resistor.

However, that's only half the battle. In order to determine how much energy that resistor has to deal with, we need to rely on Watt's law, so we can size the resistor accordingly (so it doesn't fry or explode):

Watt's Law (another 'pie' or 'circle' equation): P = IE (or I=P/E or E = I/P).

Watts = Current allowed to flow through resistor max * voltage dropped (lost) across resistor.
P = 2 * 5
P = 10 Watts.
So that 2.5-Ohm resistor has to be able to withstand or use 10 Watts. Most won't-- they'll fry because most of the resistors you're working with will only withstand 1/8th to 1/4 of a watt. Maybe 1/2. Normally you want a safety margin, so you'd normally choose a resistor that can handle twice the wattage.

Choosing the right components to control the amount of current flow and making sure you use components that can handle the power through them is a big part of the job.