공급 전압 및 전류에서 전선의 전압 강하를 어떻게 계산합니까? 최종 부하가 올바른 공급 전압을 갖도록 전압 강하를 어떻게 예상합니까?

전력 손실은 무엇입니까?

전선의 저항을 모르고 AWG ( American Wire Gauge )와 길이를 모른다면 어떻게해야 합니까?

전력 감소

다른 저항을 직렬로 배치 한 전선을보아야합니다. 이 대신, 저항 는 전압 V ...$\text{R}_{\text{load}}$$\text{V}$

저항 R 와이어 가있는 두 개의 와이어를 통해 전압 V 의 전원 공급 장치에 연결된 저항 는 다음과 같이 볼 수 있습니다 .$\text{R}_{\text{load}}$$\text{R}_{\text{wire}}$$\text{V}$

이제 V = I ⋅를 사용할 수 있습니다V는 전압, 약자 I 전류에 대한 R 저항.$\text{V} = \text{I}\cdot{}\text{R}$$\text{V}$$\text{I}$$\text{R}$

예

하자 회로에인가되는 전압이 가정 . R의 부하가 동일 250 Ω을 하고 저항 R의 와이어 입니다 2.5 Ω (당신이 와이어의 저항을 알고있다 "와이어의 저항을 계산"에서 아래를 참조하지 않는 경우). 처음에는 I = V를 사용하여 회로를 통한 전류를 계산합니다.$5\text{V}$$\text{R}_{\text{load}}$$250\Omega$$\text{R}_{\text{wire}}$$2.5\Omega$ :I=$\text{I}=\dfrac{\text{V}}{\text{R}}$$\text{I}=\dfrac{5}{250+2\cdot2.5}=\dfrac{5}{255}=0.01961\text{A}=19.61\text{mA}$

이제 우리는 한 와이어의 전압 강하가 사용하는 것을 알고 싶습니다.$\text{V}=\text{I}\cdot{}\text{R}$$\text{V}=0.01961\cdot2.5=0.049025V=49.025\text{mV}$

$\text{R}_{\text{load}}$$\text{V}=0.01961\cdot250=4.9025\text{V}$

Anticipating on voltage loss

What if we really need a voltage of $5\text{V}$ over $\text{R}_{\text{load}}$? We will have to change the voltage $\text{V}$ from the power supply so that the voltage over $\text{R}_{\text{load}}$ will become $5\text{V}$.

At first we calculate the current through $\text{R}_{\text{load}}$: $\text{I}_{\text{load}} = \dfrac{\text{V}_{\text{load}}}{\text{R}_{\text{load}}} = \dfrac{5}{250} = 0.02\text{A} = 20\text{mA}$

Since we're talking about resistances in series, the current is the same in the whole circuit. Therefore, the current the power source has to give, $\text{I}$, equals $\text{I}_{\text{load}}$. We already know the total resistance of the circuit: $\text{R} = 250 + 2\cdot2.5 = 255\Omega$. We can now calculate the needed voltage supply using $\text{V}=\text{I}\cdot{}\text{R}$: $\text{V}=0.02\cdot255=5.1\text{V}$

Power loss

What if we want to know how much power is lost in the wires? Basically, we use $\text{P}=\text{V}\cdot{}\text{I}$, where $\text{P}$ stands for power, $\text{V}$ for voltage and $\text{I}$ for current.

So the only thing we have to do is fill in the correct values in the formula.

An example

We again use the $5\text{V}$ power supply with a $250\Omega$ $\text{R}_{\text{load}}$ and two wires of $2.5\Omega$ each. The voltage drop over one piece of wire is, as calculated above, $0.049025\text{V}$. The current through the circuit was $0.01961\text{A}$.

We can now calculate the power loss in one wire: $\text{P}_{\text{wire}} = 0.049025\cdot0.01961 = 0.00096138\text{W} = 0.96138\text{mW}$

Calculating the resistance of a wire

In many cases, we will know the length of a wire $l$ and the AWG (American Wire Gauge) of the wire, but not the resistance. It's easy to calculate the resistance though.

Wikipedia has a list of AWG specifications available here, which includes the resistance per meter in Ohms per kilometer or milliOhms per meter. They also have it per kilofeet or feet.

We can calculate the resistance of the wire $\text{R}_{\text{wire}}$ by multiplying the length of the wire by the resistance per meter.

An example

We have $500\text{m}$ of a 20AWG wire. What will be the total resistance?

$\text{R}_{\text{wire}} = 0.5\text{km} \cdot 33.31\Omega/\text{km} = 16.655\Omega$