歐姆定律和計算電功率的 P= IV 是否有矛盾?

2010-05-24 2:27 am
歐姆定律和計算電功率的 P= IV 是否有矛盾?
將歐姆定律和計算電功率的 P= IV 稍微變形後,得出:
式一: I = V/R

式二: I = P/V
將兩式的V加大一倍, 式一的 I 變大一倍,式二的 I 卻變小了一倍!



註: I (電流)
V (電壓)
P(電功率)
R(電阻)

回答 (6)

2010-05-24 5:33 am
✔ 最佳答案
式一: I = V/R

if I Be the subject,it means the I be the constants,
whan a fixed Current is given,if you supplied a double voltage,the resistance will double as well to gain a congreunt current,so WHEN I IS FIXED,THE VOLTAGE AND RESISTANCE IS PROPORTIONAL.

式二: I = P/V

Similarly to above,
if I Be the subject,it means the I be the constants,
whan a fixed Current is given,if you supplied a double voltage,you gain a double power when keeping a fixed current,so WHEN I IS FIXED,THE VOLTAGE AND POWER IS PROPORTIONAL.



From the formulae,you get it

P=IV
V=IR

P=I(V)
P=I(IR)
P=I^2R
2010-05-24 8:53 am
好簡單其實
一式.V=IR
二式.P=VI

一式. V係指某兩點之間既POTENTIAL DROP. I 係指流經果兩點之間既CURRENT. R 係指兩點之間既RESISTANCE.將V變大一倍, 由於R 不變, 所以 I 會大一倍

二式.P係指某兩點之間既POWER COMSUMPTION. V 係指兩點之間既 POTENTIAL DROP. I 係指兩點之間既 CURRENT. 如果你要將V加大一倍. 由一式可見其實係要將I都會係加大一倍. 所以其實二式入面P,V,I,冇一個係CONSTANT. 若要將V加大一倍,I自然會加大一倍,最後POWER CONSUMPTION 就會變為原本的4倍

最後. P=V^2/R
根據頭先所講, 你將V加大一倍, POWER應該係變原本4倍,
呢條式入面RESISTANCE OF LOAD 係唔會變, P=4(V' ^2)/ R
所以呢條式係岩既

至於呢條式點黎, 咪就係結合式一同式二, 我諗你都係因為覺得式一式二錯先問呢個問題
2010-05-24 7:13 am
式一同式二係兩回事

式一: I = V/R 計電阻, 電流與電壓

式二: I = P/V 計功率, 電流與電壓

式一非常簡單, 電壓除電阻等於電流, 但係冇計功率, 代替計算就係
V = IR
I = V/R
R = V/I

但係式二就複雜啲, 電壓乘電流等於功率係冇錯, 但係如果計電壓電阻與功率就麻煩啲, 方法有2

1. 先計 P = VI, 然後 R = V/I
2. 直接計 P = V^2/R 或者 P = I^2R

擧個例子

V = 3.6V
I = 2.4A
R = 1.5Ω
P = 8.64W


V = IR = 2.4A * 1.5Ω = 3.6V
I = V/R = 3.6V / 1.5Ω = 2.4A
R = V/I = 3.6V / 2.4A = 1.5Ω

P = VI = 3.6V * 2.4A = 8.64W

P = V^2/R = 3.6V * 3.6V / 1.5Ω
= 12.96 /1.5Ω = 8.64W

P = I^2R = 2.4A * 2.4A * 1.5Ω
= 5.76 * 1.5Ω = 8.64W

希望幫到你


2010-05-24 4:25 am
Your equation (1) I = V/R is a basic law, the Ohm's Law. It is true under the condition that the resistance R is constant. Hence, any increase of voltage V leads to a corresponding increase of current I.

In equation (2), current I and voltage V are inversely proportional if, and only if, the value of power P is a constant. Under normal circumstances, this is not the case. From Ohm's Law, power P can be expressed as V^2/R. Hence, your equation (2) is actually, I = [V^2/R]/V

When V increases, the power P not only increases, but would increase to a much larger extend because P depends on V^2. As a rsult, this leads to the increase in current I as calculated from equation (1).

You could easily see that equation (2) and equation (1) are actually identical. I = = [V^2/R]/V just reduces to I = V/R
2010-05-24 2:51 am
唔知你認為有咩矛盾呢﹖式一中I同V係正比,而式二I同V係反比,只要你兩條式係正確嘅話,睇唔到有問題。或者你去觀察一啲電器可以用110V同220V,你睇睇灰士(FUSE/保險絲)係110V或220V時應該用幾多A,可以話你知係唔同AMP數,你想知點解,用你嘅公式計計。
2010-05-24 2:45 am
問題在於式二,若V加大一倍, P都會加大一倍,而不是I 變小了一倍。
此外歐姆定律和計算電功率的 P= IV 是沒有矛盾,因P=IV, (I square)R=IV, V=IR。它們都是相通的。
參考: my knowledge


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