将“6V 3W”的灯泡L1和“6V 2W”的灯泡L2串联后接入6V电路中,L1的实际电功率为 W,并联后接入6V电路中,电路的总功率为 W.
【答案】
分析:(1)由题意知,两灯泡的电阻不变,电源的电压不变,根据灯泡的额定电压和额定功率以及电功率的公式P=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104182551802059322/SYS201311041825518020593002_DA/0.png)
,R=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104182551802059322/SYS201311041825518020593002_DA/1.png)
求出灯泡的电阻.
(2)两灯串联时,根据串联电路的特点,求出通过灯泡L
1的电流,再用公式P=I
2R,求出L
1的实际功率.
(3)当两灯并联时,它们均工作在额定电压下,各自消耗的实际功率即为额定功率,相加即为电路总功率.
解答:解:(1)两个灯泡的额定电压和额定功率分别是U
1额=6V,U
2额=6V,P
1额=3W,P
2额=2W,
R
1=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104182551802059322/SYS201311041825518020593002_DA/2.png)
=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104182551802059322/SYS201311041825518020593002_DA/3.png)
=12Ω,R
2=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104182551802059322/SYS201311041825518020593002_DA/4.png)
=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104182551802059322/SYS201311041825518020593002_DA/5.png)
=18Ω;
(2)灯泡L
1、L
2串联,电路中的电流I=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104182551802059322/SYS201311041825518020593002_DA/6.png)
=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104182551802059322/SYS201311041825518020593002_DA/7.png)
=
![](http://thumb.1010pic.com/pic6/res/czwl/web/STSource/20131104182551802059322/SYS201311041825518020593002_DA/8.png)
=0.2A,
I
1=I=0.2A,P
1=I
12R
1=(0.2A)
2×12Ω=0.48W;
(3)灯泡L
1、L
2并联时,加在两灯上的电压均为6V,两灯都能正常发光,所以消耗的电功率为:
P
1实=P
1额=3W,P
2实=P
2额=2W.
所以电路总功率为:P
总=P
1实+P
2实=3W+2W=5W.
故答案为:0.48W;5W.
点评:根据额定电压和额定功率求出电阻,并且利用求出的电阻计算其他物理量,这是很常用的方法.