我想問 物理pastpaper零六年電學LQ一題

I don't have the question in hand. But from reading the part that you posted, have you noticed the key words...IF THE INTERNAL RESISTANCE OF THE DRY CELLS IN NOT NEGLIGIBLE... .?

When R (I presume that this refers to an rheostat) is reduced, the current in the circuit increase. As such, the potential drop across the internal resistance of the dry cells (i.e. the lost voltage) will also increase. As a result, the voltage across the bulb is lower than before because of voltage sharing with the internal resistance of the cells. The bulb thus gives less brightless.

因為 power supply 可供應的輸出電流足夠燈泡所需, 因此有足夠光度, 但若用電池, 它本身的電流輸出量是有限制的, 這大小是視乎電池內阻的大小.

因此當用電池做實驗時, 燈泡較暗的原因是, 燈泡所需正常光度的電流超過電池可正常提供的電流.

電池的內阻會使電池的端電壓下降, 燈泡用的電流愈大, 電池內阻做成的電壓降亦越大, 而電池的電動勢 1.5V 是固定的. 若電池內阻做成的電壓假設是0.4V, 那麼電池的端電壓便降低至 1.1V. 若燈泡所需的電流很小, 內阻的電壓降假設 0.02V, 那麼電池的端電壓仍有 1.48V.

電池的端電壓=電池的電動勢-內阻做成的電壓降(Ir)

因此電池的端電壓(即電池正負極), 是會隨著負載 R 而變化的, R 越大(電流小), 電池的端電壓越接近電池內部可產生的電動勢. R 越小(電流大), 電池的端電壓越降低. 因此 R(燈泡)便得不到正常電壓, 而轉暗.

你問題中所指, when R is reduced, what R ?????是不是指燈泡.

buld的光暗取決於current 大小

by ohm's law, v = ir , 即係 i=v/r

相同的voltage, r愈大, i愈小
r愈小, i愈大, buld愈光, 明唔明

2007-03-12 23:27:58 補充：
sorry, 原來the brightness of the bulb decreases when R is reduced咁我真係唔明, power dessipated by bulb= iv = v*v/R = light energe of bulb, 點計都係R愈小，愈光

2007-03-14 00:06:15 補充：
下面那個人點醒了我Let the internal resistance of bulb = ri = V/(r R)power dissipated by bulb = i*iR = R*i^2=RV^2/(r R)^2

2007-03-14 00:06:39 補充：
(Continue)R/(r R) infinity, R/(r R) = 1, the larger is the power dissipated by bulbThe Smaller R, the smaller R/(r R) e.g. if R = 0 , R/(r R) = 0, power = 0, less bright.