U = R * I – A little Intro

Hi, I had thought, since I had done nothing with electronics for a long time, to make a small post about the basics of electrical engineering.

Electrical engineering is the one with the wires and lights and chips. Yup that's it in the rough. Well, there is a bit more to it than wires and lamps. Actually, you develop circuits and calculations for it, mainly Ohm's law.

Let's consider this highly complex circuit as a small example.

We see here a 5V battery, a resistor $R1$ and a LED $D1$, which are connected together. Together they form a closed circuit. But what does this have to do with Ohm's law?

Ohm's law

Well, Ohm's law describes the relationship between voltage and current and the resistance.

The formula is then $U = RI$.

$U$ is the voltage in V
$R$ is the resistance in ohms $\Omega$
$I$ is the current in amperes (A)

Ok, What is voltage in principle? Voltage is a potential difference between two poles. Voltage always has a balancing tendency. This also indicates how strong the drive of the current is.

The current, is the flow of electrons. This is flowing in the technical representation from plus to minus. In practice, however, it is exactly the opposite.

The resistor, well the resistor just. No joking aside, a resistor is also seen as a consumer.

Examples

Suppose we have a $100\Omega$ resistor with 50mA flowing through it. What is the voltage that drops across the resistor?

$I = 50 mA$
$R = 100 \Omega$

$U = RI$
$U = 100 \Omega * 50mA$
$U = 5V$

The voltage is therefore 5V, which drops at the resistor.

How does it look now, if we leave the resistance the same and halve the current?

Exactly the voltage is halved $U = 2.5V$. What does this mean? The resistance is constant, because it is mostly a physical component.

But what happens to the current if we put in a $200 \Omega$ resistor, and leave everything else?

$U = 5V$
$R = 200 \Omega$

$I = \frac{U}{R}$

$I = \frac{5V}{200 \Omega}$

$I = 0.025 \Leftrightarrow 25mA$

The current is halved, and you can double it again by halving the resistance. Here you can say the higher the resistance, the higher the voltage that falls on it. With the current, however, vice versa.

Ok, let's calculate the circuit from above. A few values are already known.

Given: $I = 20mA$
$U_{D1} = 1.6V$

Searched: $U_{R1}, R1$

$U_{R1} = U - U_{D1} \Leftrightarrow 5V - 1.6V \Leftrightarrow 3.4V$

$R1 = \frac{U_{R1}}{I} \Leftrightarrow \frac{3.4V}{20mA} \Leftrightarrow 170 \Omega$

$R_{D1} = \frac{U_{D1}}{I} \Leftrightarrow \frac{1.6V}{20mA} \Leftrightarrow 80 Ohm$

That was simple. But why had the current not changed? Quite simply, since this is a series circuit, the current is the same everywhere.

In a parallel circuit, the voltage is the same everywhere, but the current is not. A small extension of the complex circuit to get a parallel circuit looks like this.

What do we have here exactly? In principle, you can divide circuits into smaller circuits. You look at the whole circuit, then in small how do you say bites. I have simply drawn the voltage drops (red) to show a little bit how you can divide the circuit.

We know from just now that the voltage is the same in a parallel circuit. We can now take advantage of this. We consider the strings with $R1$, $D1$ and $R2$ as separate series circuits. Let's calculate that.

The resistor $R1$ has $170 \Omega$ and a voltage drop of 3.4V, as just and the LED $D1 = 1.6V$ voltage. The total current Ig is 30mA and the total voltage $U_{0} = 5V$.

Then we can calculate:

$U_{1} = 3.4V + 1.6V \Leftrightarrow 5V$

$I_{2} = I_{0} – I_{R1} \Leftrightarrow 30mA – 20mA \Leftrightarrow 10mA.$

$R_{2} = \frac{U_{0}}{I_{2}} \Leftrightarrow \frac{5V}{10mA} \Leftrightarrow 500 \Omega$

$I_{0} \Leftrightarrow I_{1} + I_{2} \Leftrightarrow 20mA + 10mA \Leftrightarrow 30mA$

$U_{0} \Leftrightarrow U_{1} \Leftrightarrow U_{2}$

$R_{t1} = R_{1} + R_{D1} \Leftrightarrow 170 \Omega + 80 \Omega \Leftrightarrow 250 \Omega$

$R_{t2} \Leftrightarrow 500\Omega$

$R_{0} = \frac{1}{\frac{1}{R_{t1}} + \frac{1}{R_t2}} \Leftrightarrow \frac{1}{\frac{1}{250\Omega} + \frac{1}{500\Omega}} \Leftrightarrow \frac{1}{0.004 + 0.002} \Leftrightarrow \frac{1}{0.006} \Leftrightarrow 166.66 \Omega$

Here we see that the total resistance is not larger, as we would have expected, but the total resistance is smaller than the smallest individual resistance. The formula to calculate the total resistance of a parallel circuit also works for $n$ individual circuits.

How to measure current and voltage as well as resistance? The measurements are relatively simple. A small sketch shows how to connect the measuring instruments in the circuit.

We see here that the voltmeter (V) is connected in parallel with the resistor. With this, we can now measure how much voltage drops across the resistor. Well here the total voltage. To measure the resistance, we also connect our meter (R) directly in parallel with the resistor. The current, however, we can only measure, if our measuring device (A /Ampermeter) is connected in series, as you can see on the picture. In some cases, in order to measure the current, it may be necessary to cut the affected wires and connect the device in between.

I hope I was able to give a little insight into electrical engineering and the Ohm's law 😄