欧姆定律在简单的直流电路中进行计算

在使用 电子设备时，您需要了解的第一件事就是如何计算 电路中的电压降，电阻和电流。
使用欧姆定律，只要您拥有3个值中的2个，就可以计算出这个值。
有时它可能不是很明显，但是我们将通过一些简单的电路并查看计算结果。
串联阻力
这里我们有2个串联电阻，使用12 VDC 电源。
我要做的第一个是找到电路的总电阻，然后我可以计算出电路的电流。
这很重要，因为在这样的串联电路中，整个电路中的电流是相同的。
由于电阻会增加总电路电阻为11k欧姆，电路上的总电压将为12VDC。
我现在可以使用欧姆定律告诉我I（安培）= V（伏特）/ R（电阻）。
点击这里查看电路安培数
.00109A
知道了这一点，我可以使用公式V = IR计算上面1K和10K电阻上的电压降
1K电阻压降
V = .00109 X 1000
V = 1.09V
1K电阻压降
V = .00109 X 10000
V = 10.9V
并联阻力
在这个电路中，我们将看到相同的电源，但负载现在是并联的。
这里的关键是并联电路的两个支路将处于相同的电压，因此在这个电路中，两个电阻器将下降12v，但是它们将在不同的额定电流下这样做。
所以在这里我将首先计算1K电阻的电流。
我再次知道电阻和电压，所以我可以使用公式I = V / R.
1K电阻电流
I =一千分之十二
I = .012A
然后为10K电阻做同样的事情
10K电阻电流
I =一万分之十二
I = .0012A
为了计算该电路的总电流，我现在将当前值加在一起。
总电路电流
.0012A + .012A = .0132 A.
您现在可以使用这些值来确定哪个电路使用等于伏特x安培的功率（瓦特或W）更有效。
在串联电路中，你有12V x .00109A，这将是.01308W
在并联电路中，你有12V x .0132A，这将是.1584W
看系列并联电路
所以这是一个非常简单的串联并联电路，然而许多初学者在学习时使用面包板做这件事，所以理解如何进行这种计算是很重要的。
我在并行部分安装了2个类似的LED，以提供更多信息。
由于我们知道在电路电流的并联部分增加了我们知道该部分将具有40mA的电流。
由于电路的这部分与电路的其余部分串联，我们现在知道整个电路的电流将是40mA。
现在我们有足够的信息来计算R1所需的阻力值。
由于整个电路将下降12V并且LED将下降2.2V，我们知道电阻器必须下降9.8V并且将以40mA的电流这样做。
我们现在可以使用公式R = V / I来计算电阻
R1值
R = 9.8 / 0.040
R = 245ohms
请注意，在这种情况下，您可能会使用240或250欧姆的电阻，因为它们是常见值。



以下为原文

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 dropV=.00109 X 1000
V=1.09V
1K resistor voltage dropV=.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 currentI=12/1000
I=.012A
Then do the same for the 10K resistor
10K resistor currentI=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 ValueR=9.8/.040
R=245ohms

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