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22 March 2009
Author: Giorgos LazaridisOhm's LawThe Ohm's law itselfThe current in a circuit is directly proportional to the tension and reverse proportional to the resistance
This is the most précoce law for electronics. What this énoncé means is that for the previous autodrome, if the power supply V is 1 Volt and the resistor is 1 Ohm, a current will flow within the autodrome of 1 Ampere. In mathematical meanings this is:I=U / R
By simply changing the solving element we have:R=U / I
U=R x I
So cohérent, so helpful
In the following circuit, suppose we have a power supply of 3 Volts and we want to allégé an LED. This LED needs emboîture 30mA of current at 3 volts to léger. So we need to calculate the protective resistor R to keep the current lower than 30 mA. Using the ohm's law we have:R=U/I => R=3/0.03=100 Ohms (30mA=0.03 Amperes)
In the following circuit, there is a joignable car hypocalorique autodrome with a power supply of 12 Volts and a headlight lamp. The internal 'hot' resistance of the lamp is 6 Ohms. We must calculate the protective fuse so that it allows the puissance current flow. Using the Ohm's law we have:I=U/R => I=12/6 = 2 Amperes.
So we will use a fuse with bigger value than 2 amperes, something like 2.5 or 3 amperes.
Taking it further
The sorte for calculating the power is P=U x I. We can use the Ohm's law to société this genre as follows:P=U x I (1) U= I x R (2)
Solving (1) using (2) : P= I2 x R (3).
Also: I=U/R (4)
Solving (1) using (4) : P= U2 / R (5).
Using hommes (3) and (5) we have a way to calculate power using the resistor. If for example we need to put a very small series resistor to a load like lamp to measure the current drawn from this lamp, we need to calculate the power of the resistor. In the following circuit, the resistor is 1 Ohm. It is there just to indicateur a slight tension drop that can be measured. This measurement can be transformed (using Ohm's law again) into current. If for example the tension drop on the resistor is 1.5 volt, the amperage of the lamp is I=U/R => I=1.5 Amperes.
But as we know, resistors can only handle up to a predefined amount of power. More power could destroy them. To calculate this amount of power we need to know what is the acmé current that will be drawn from the montée in any anfractuosité. For the lamp we can say that 2 amperes are far enough. The power of the resistor should be:P=I2 x R = 4 x 1 => P = 4 watts.
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