In my last post, I introduced the basic quantities used in electronics. These notions are necessary to move further with electrical circuits calculation. It is easier to learn the Ohm's law with examples so here we go :

The picture above is what we call a circuit schematic. At the left of the schematic we have the DC source (batteries in this example). At the right you see the resistor schematic symbol. Keep in mind that the wire at the bottom is the ground net. We have a voltage source of

**10** Volts and a resistance of

**5** Ohms. We want to find the current that the supply will source through the resistor. If we recall that

**E = R x I **we can solve for the current

**I = E/R **then

**10/5 = 2 **Amps. But what if we've got multiple resistors?

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Resistors in series

If you have multiple resistors in series, you just have to sum their values and replace all the resistors by one resistor that is the total of the others. See this circuit :

The 2 and 3 Ohms resistors can be replaced by a resistor of 5 Ohms. You then are **I = E/R**, **5/5 = 1 ** Amps.

Remember : what we are doing here is circuit reduction. In the above example, we reduced the circuit to find the total current the goes through the resistors. In a series circuit, **the current is the same through the series component.** This means that above, you have **1** Amps through **R1** **and R2**. To completely solve the circuit, you would want to know the voltage accross the two resistors. Easy! Remember **E = R x I **? For **R1**, **E = 2 x 1 = 2 **Volts. I hope you guess that we have **3 **Volts across **R2**.

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Resistors in parallel

For parallel resistors, the formula is a little bit more complicated. When we have two resistors **R1 **and **R2**, the parallel resistance **R3 **is calculated with **1/R3 = 1/R1 + 1/R2**. You can reduce the formula to **R3 = (R1 x R2) / (R1 + R2)**. After a while, you will realize that if two resistors of the same value are in parallel, the result will be the half of the resisor value. Proof : **(R0 * R0) / (R0 + R0) = (R0^2) / (2*R0) = R0/2**, half of **R0**.

In the above circuit, the resistors are the same, **20** Ohms. **R3 = (R1 * R2) / (R1 + R2) = ( 20 * 20) / (20 + 20) = 400 / 40 = 10** Ohms. The current sourced by the batteries : **I = E/R**, **24/10 = 2.4 **Amps.

Observe that the resulting resistor is always at a lower value than the smaller of the two resistors value.

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Units in electronics

In the circuits above, we have several amps through the resistors. This means a lot of heat and big resistors (physically). Most of the time, in electronics circuits, we have currents about 1000 times smaller. The resistors used are several thousands times bigger than the ones used in the examples.

The most common SI (International System) units are : milli (1/1000) and kilo (1000x). Generally, we have milliamps, kiloohms, and volts stay the same. If you multiply kiloohms and milliamps, you end up with volts. The fun part is that you can calculate everything without caring too much about the units, just remember that you have milliamps and kiloohms. Example (from above) : **R = 5 **kiloohms, **E = 10 **Volts : **I = E/R **; **10/5 = 2 **milliamps. See? The units are not the same but the numbers are. Don't worry, it will eventually sink into your brain...

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Shortcut : voltage dividers

When you have 2 resistors in series (see example above) you can easily find the voltage across each resistors. This is needed to completely solve the circuit. **V2 **is going to be the voltage across **R2**. If **R2 (3**Ω**)** is next to **GND** (ground) and **R1 ****(2**Ω**)**** **is** **next to the positive supply : **V2 = E * (R2 / (R1 + R2))**. In the example of the series resistors : **V2 = 5 * (3 / (2 + 3)) = 5 * (3 / 5) = 15 / 5 = 3**. We then have **3 **Volts across **R2**, exactly what we found earlier. Here is a little form to test voltage dividers :

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Conclusion

Okay, I have to admit that a lot of boring calculations were done in this post, but they are needed for the most simple circuits and this is the basis to understand all the later examples. This will help us to get lower voltages from a higher voltage source with the right combination of resistors. We will also be able the verify the current flowing through our components and be sure that the current ratings are respected.

The next post will quickly cover the operation of the diode, and we will see how to light some LEDs (Light Emitting Diodes). This will be a lot of fun =D