One of the first things that you you need to know when working with electronics is how to calculate voltage drop, resistance, and current in a circuit. Using ohms law you are able to calculate this as long as you have and 2 of the 3 values. At times it may not be obvious, however we will go through a few simple circuits and look at the calculations.

#### Resistance in Series

Here we have 2 resistors in series using a 12 VDC power supply. The first this I am going to do is find the total resistance for the circuit and then I can calculate the current for the circuit. This is important because in a series circuit like this the current will be the same throughout the entire circuit.

Since resistance will add the total circuit resistance will be 11k ohm and the total voltage dropped over the circuit will be 12VDC. I can now use ohms law which tells me that I(amps)= V(volts) / R (Resistance).

## Click Here for circuit amperage

.00109A

Knowing this I can calculate the voltage drop across the 1K and 10K resistors above using the formula V=IR

## 1K resistor voltage drop

V=.00109 X 1000

V=1.09V

## 1K resistor voltage drop

V=.00109 X 10000

V=10.9V

#### Resistance in Parallel

In this circuit we will look at the same power supply, however the loads are now in parallel. The key here will be that both legs of the parallel circuit will be at the same voltage, so in this circuit both resistors will drop 12v, however they will do so at different current ratings.

So here I will start by calculating the current for the 1K resistor. Again I know the resistance and the voltage so I can use the formula I=V/R.

## 1K resistor current

I=12/1000

I=.012A

Then do the same for the 10K resistor

## 10K resistor current

I=12/10000

I=.0012A

To calculate the total current for this circuit I will now add the current values together.

## Total Circuit Current

.0012A + .012A = .0132 A

You could now use these values to determine which circuit was more efficient using the power (in Watts or W) which is equal to Volts x Amps.

In the Series circuit you have 12V x .00109A which would be .01308W

In the Parallel circuit you have 12V x .0132A which would be .1584W

#### Looking at a Series Parallel Circuit

So this is a very simple series parallel circuit , however something many beginners do this with a breadboard while learning so it is important to understand how to do this calculation.

I have installed 2 similar LEDs in the parallel portion to give some more information. Since we know in the parallel portion of the circuit current adds we know that portion will have a current of 40mA. since this part of the circuit is in series with the rest of the circuit we now know that the current for the entire circuit will be 40mA.

Now we have enough information to calculate the value of the resistance needed for R1. Since the entire circuit will drop 12V and the LEDs will drop 2.2V we know the resistor will have to drop 9.8V and will do so with a current of 40mA. We can now calculate the resistance using the formula R=V/I

## R1 Value

R=9.8/.040

R=245ohms

Note in this instance you would likely use a 240 or 250 ohm resistor as they are common values.