当在电阻r上应用一些电势差V，则电场在电荷q上完成的工作在时间t上流过电路的工作将为w = qv = vit = v = v = v vit = v2T/4.2R焦耳。这项工作似乎是电阻器中的热能。
Heat produced by the resistance R is H = W/J = V2T/4.2R CAL。这种关系称为焦耳加热。
电力
The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.
P = W/t = Vi =i2r=V2/r
Units:It's S.I. unit is Joule/sec or Watt
更大的S.I.单位是KW，MW和HP，请记住1 HP = 746瓦
额定值
在电器上（灯泡，加热器，间歇泉...等）。瓦数，电压等命名为额定值，例如如果考虑到我们的光为40 W，220 V，则额定功率P = 40 W，而印刷电压V = 220V。
电器的抗性
如果忽略了用温度的电阻变化，则可以通过额定功率和额定电压来计算任何电器的电阻，即通过使用r=V2/pr
电力消耗（照明）
一种n electrical appliance (Bulb, heater, ... etc.) consume rated power (Pr) only if applied voltage (V一种）等于额定电压（vr），即如果v一种= Vr
所以p = v2/ R也r=V2/pr。
Long distance power transmission
When power is transmitted through a power line of resistance R, heat-loss will be i2r
Now if the power P is transmitted at voltage V then P = Vi , i.e. i = (P / V)
So Power loss = P2/v2*R
现在，对于给定的线和功率，R和P是恒定的，因此功率损失
So if power is transferred at high voltage, power loss will be low and vice-versa. That is why long distance power retransmission is carried out at high voltage.
1。The price of electricity consumed is calculated on the basis of electrical energy and not on the basis of electrical power.
2。The unit Joule for energy is very small hence a big practical unit is considered known as kilowatt hour (KWH) or board of given unit (B.T.U.) or simple unit.
3. 1 KWH or 1 units is the quantity of electrical energy which dissipates in one hour in an electrical circuit when the electrical power in the circuit is 1 KW thus 3.6*108J.
4. Important formulae to determine the no. of used units is
Series combination
(i) Total power consumed 1/Ptotal= 1/P1+1/P2... (ii) If 'n' bulbs are identical, Ptotal= P/N (iii) Pconumed∝ V ∝ R ∝ 1/Pratedi.e. in series combination bulb of lesser wattage will give more white light and p.d. seem across it will be more. |
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平行组合 （i）总功率消耗ptotal= P1+P2。。 (ii) If 'n' identical bulbs are in parallel Ptotal= np (iii) Pconumed∝1/r ∝ pratedI.I.E.在平行的组合中，较大的瓦数将产生更多的白光，并且更多的电流将通过它。 |
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