将“6V 3W”的灯泡L1和“6V 2W”的灯泡L2串联后接入6V的电源,实际电功率之比为______,并联后仍接入6V电源中,通过的电流之比为______,此时电路的总功率为______W.
【答案】
分析:(1)由灯泡的铭牌可知,灯泡的额定电压和额定功率,根据R=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104180737285358835/SYS201311041807372853588009_DA/0.png)
求出两灯泡的电阻;串联时根据P=I
2R可求实际电功率之比;
(2)并联后仍接入6V电源中,因灯泡能正常工作,则根据I=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104180737285358835/SYS201311041807372853588009_DA/1.png)
求出通过的电流之比,此时电路的总功率利用P=P
1额+P
2额求得.
解答:解:灯泡L
1电阻R
1=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104180737285358835/SYS201311041807372853588009_DA/2.png)
=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104180737285358835/SYS201311041807372853588009_DA/3.png)
=12Ω,灯泡L
2的电阻R
2=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104180737285358835/SYS201311041807372853588009_DA/4.png)
=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104180737285358835/SYS201311041807372853588009_DA/5.png)
=18Ω,
(1)串联后接入6V的电源,通过两灯泡的电流相等,
则根据P=I
2R可得:
实际电功率之比:P
1:P
2=I
2R
1:I
2R
2=R
1:R
2=12Ω:18Ω=2:3;
(2)将它们并联后仍接入6V电源中,灯泡的实际电压和额定电压相等,灯泡能正常工作,
所以,通过两灯泡电流的比为I
1:I
2=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104180737285358835/SYS201311041807372853588009_DA/6.png)
:
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104180737285358835/SYS201311041807372853588009_DA/7.png)
=P
1额:P
2额=3W:2W=3:2;
P=P
1额+P
2额=3W+2W=5W.
故答案为:2:3;3:2;5.
点评:本题考查了串联电路和并联电路的特点,欧姆定律、电功率公式的计算;关键是相关公式和规律的灵活运用,以及理解和掌握灯泡铭牌的含义.