phy-電路

Assume the supply voltage remains the same throughout.

If an additional light bulb is connected in series to another one,
total resistance increases, current decreases, and the bulbs glow dimmer.
Because the current through each bulb decreases, the voltage across each bulb also decreases.

In a parallel connection, the voltages across each bulb are the same (which equals to the supply voltage). Thus, the current through each bulb remains the same, but the total current increases. The brightness of the bulb is unchanged (because current through it remains the same as before).
The reverse happens when you reduce the number of bulbs from two to one.

in SERIES WIRING

the loading increases,the TOTAL RESISTACNE increases,
it cause the TOTAL CURRENT decreases,
and therefore the brightness will also decreases
and the TOTAL VOLTAGE remain unchange.


in PARALLRL WIRING

the loading increases,the TOTAL RESISTACNE decreases,
it cause the TOTAL CURRENT increases,
and therefore the brightness will also increases
and the TOTAL VOLTAGE remain unchange.

在串聯中, 由本身一個燈泡增至兩個燈泡, total resistance會大左 = 對

所以燈泡會光左? = 錯

咁total I 同total V 有咩變化? = total I = 減 同total V = 冇變

如情況相反(兩個燈泡減至一個),
total R、I、V同光度又有咩變化?
total R = 減 , I = 加 , V = 冇變 , 同光度 = 加



在並聯中, 由本身一個燈泡增至兩個燈泡, total resistance會小左 = 對

所以燈泡會光左? = 對

咁total I同total V有咩變化?
total I = 加倍 同total V = 冇變

如情況相反(兩個燈泡減至一個),
total R、I、V同光度又有咩變化?
total R = 加倍 , I = 加 V = 冇變 , 同光度 = 減半

係串聯電路，兩個燈泡未必會光咗，你嘅講法係兩個燈泡光過一個燈泡嗎，事實係似是而非。

由於電阻增大，電流減少，燈泡電功率亦下降，假設兩個燈泡電阻電壓功率相同，一個燈泡功率係I^2R，兩個燈泡串聯時功率係2(I/2)^2R=I^2/2R，即係一個燈泡時一半功率，所以應該會暗咗。

並聯時，電阻下降，電流增加，總功率係(2I)^2(R/2)=2I^2R，係一個燈泡時兩倍功率，所以係光咗，但要留意燈泡有可能超過負荷而燒燬。