Easy to remember triangle. Electricity, power

[ttwhin]

Maybe one can find this posted before, didn't check

For those who don't remember and for beginners.
If one want to try to calculate some around electricity here are some easy to remember formula in triangle from Ohms law

For calculate the one unknown, from two known.

V= Electrical force measured in Volt
I = Current measured in Ampere
R= Resistance measured in Ohm

../ \
./ V \
/ IxR\

From top divide with one under to find answer for 3. unknown. From under multiply on same line. Exampled V= IxR or I=V/R



To calculate to/from power

P= Power measured in Watt


..../\
../ P \
/ VxI \




Exampled VxI=P.
3V x 8A = 20Watt.

Power work-hour
If one have one known factor of time, one can calculate also work-hour.
Just put numbers in

Exampled: One know current-hour (Ampere hour) is 5Ah and Volt is 8,4V.
8,4V x 5Ah = 42Wh (42 watt hour)

Remember this triangles is easy !

btw i am no teacher and will never be, so that is said

If anyone find error in my in-hurry written lesson, post on.
i try to correct

 

[jhellwig]

Here is a simple one:

On the simple one you are supposed to put your finger on what you are trying to find and do the calculation.


Your triangle is a good idea but someone beat you to the punch. Every electrical theory class, book, what have you uses the circle thingie.

 

[MorePower]

Here is a simple one:

On the simple one you are supposed to put your finger on what you are trying to find and do the calculation.


Your triangle is a good idea but someone beat you to the punch. Every electrical theory class, book, what have you uses the circle thingie.

That's great, except the caption for your second picture is wrong, because it's E=I*R, not E=I/R.

 

[cckw]

yeah, and Pie - R - squared

 

[ttwhin]

Your triangle is a good idea but someone beat you to the punch. Every electrical theory class, book, what have you uses the circle thingie.

Of course they have
I`m not trying to beat that, its a repeating for those who don't remember and for beginners. Perhaps i should stated that in first post.
Its not a competition here you know
Ok, i edit

Btw now i seen the third symbol for electrical force: E, before it was used U, and i checked on Wikip, they used V. So i thought its was new symbol for Volt and used in my post.

 

[jhellwig]

That's great, except the caption for your second picture is wrong, because it's E=I*R, not E=I/R.

I didn't even notice that. It was just some random one off of google.


ttwhim
I wasn't trying to compete with you. You got a good idea but it isn't much different than the circle. If you tried teaching someone with your triangle they would probably get confused. I learned using the circle when I was a "beginner". I think the instructor even related it to pie. Take the slice out that you want to find. Then divide or multiply the rest. The number you are looking for should be the size of the slice you took out. Big slice=big number. Small slice=small number.

I am not putting you down but you are trying to reinvent the wheel.

 

[kosPap]

my approach is a bit different....

EIR

or E=I*R

then I do the math

 

[fyrstormer]

Calculating the resistance necessary to drop input voltage down to what I need has always baffled me. The rules W=VI and I=V/R are a piece of cake, though.

 

[Selectron]

Calculating the resistance necessary to drop input voltage down to what I need has always baffled me. The rules W=VI and I=V/R are a piece of cake, though.

I assume you mean in a basic series circuit where the supply voltage is too high for the load (e.g. LED, fan, relay coil etc.) where you want to drop the voltage difference across a resistor. If so then that's very straightforward provided that you know how much current the load will draw. Just take the supply voltage, subtract the voltage across the load, and then divide the result by the current which will flow in the circuit (in amps) - the result is the value of the resistor in ohms.

As an example, suppose we want to feed a standard red LED from a 6 volt supply. We know that a red LED has a forward voltage drop (Vf) of around 1.8 volts, and maximum current is around 30mA but they're typically fed with 20mA or less, so let's say we want to run it at 15mA (0.015A).

So, we have a supply of 6 volts, the voltage across the LED will be 1.8 volts, so we know that we need to drop 4.2 volts across the resistor - that's 6 - 1.8 = 4.2. We also know that the current which will flow in the circuit will be 15mA.

We know that V=I*R, so therefore R must equal V/I, so if we add some numbers we get:

R = V / I
R = 4.2V / 0.015A
R = 280 ohms

Before we select the resistor we should consider how much power it will dissipate, so we can use P = I*V, where P is the power in watts, I is the current through the resistor in amps, and V is the voltage across the resistor in volts, so:

P = I * V
P = 0.015 * 4.2
P = 0.063 watts (63mW)

So if we were using the E24 (5%) resistor series, which doesn't contain a 280 ohm resistor, we would choose the next highest value, which would be 300 ohms and since it will only dissipate less than a tenth of a watt, a component rated at one quarter watt would be plenty.

Now that we know we'll be using a 300 ohm resistor, we can quickly run through the calculations again to determine just how much current will flow. We still have Vsupply minus Vload = 6 - 1.8 = 4.2 so 4.2V is still the voltage across the resistor, and if we now use I=V/R then we have 4.2V / 300 ohms = 0.014A

So 14mA will flow in our circuit, and it will look like this: