(2011?常德)如图所示,一个白炽灯泡L标有“6V 6W”的字样,现将它与一定值电阻R并联后接入电源电压恒

2024-11-26 22:40:33
推荐回答(1个)
回答1:

(1)灯泡的电阻为RL=

U2
P
=
(6V)2
6W
=6Ω;
(2)灯泡的额定电流为IL=
P额
U额
=
6W
6V
=1A,
通过电阻的电流为IR=I-IL=1.5A-1A=0.5A,
定值电阻的阻值为R=
U
IR
=
6V
0.5A
=12Ω;
(3)定值电阻R消耗的电功率为PR=UIR=6V×0.5A=3W.
答:灯泡L的电阻值为6Ω;定值电阻R的阻值为12Ω;定值电阻R消耗的电功率为3W.