题目内容
将标有“6V 3W”的灯泡接在3V的电路中时,灯泡消耗的功率是 W.若要将其接入9V的电路中并使它正常发光则应串联一个阻值是 Ω的电阻.
R=
=
=12Ω,P=
=
=0.75W;
I=
=
=0.5A,R串=
=6Ω;
故答案为:0.75W,6Ω.
| U2 |
| P |
| (6V)2 |
| 3W |
| U2 |
| R |
| (3V)2 |
| 12Ω |
I=
| U |
| R |
| 6V |
| 12Ω |
| 9V-6V |
| 0.5A |
故答案为:0.75W,6Ω.
练习册系列答案
相关题目