小刘想知道小灯的亮暗程度与什么因素有关.于是找来额定电流均小于0.6A,额定电压是2.5V的灯L1和额定电压

2025年03月16日 19:55
有1个网友回答
网友(1):

(1)L1和L2的额定电流均小于0.6A,电流表选择0~0.6A,电流表串联在电路中.
滑动变阻器已经选择下面接线柱,上面任意选择一个接线柱,滑动变阻器串联在电路中.
所以,电流表的“0.6”接线柱与滑动变阻C(或D)相连,电压表的“-”接线柱与灯泡左端相连.
如下图:

(2)滑动变阻器接入A接线柱,滑片在B端,滑动变阻器连入电路电阻最大,电流最小,起到保护电路作用.继续实验时,滑动变阻器阻值减小,滑片向左端移动.
灯泡亮度跟实际功率有关,实际功率P=UI,需要记录电压和电流,电流的单位为A.
(3)从实验1→实验2→实验3,灯泡越来越亮,灯泡实际功率越大.
(4)L1的额定功率为2.5V,电压表选择0~3V量程.
L2的额定功率为 3.8V,电压表选择0~15V量程.
所以L2替换L1,同时更换电压表量程.
(5)灯丝电阻受温度影响较大,灯泡在不同电压下,温度不同,灯丝中电阻不同,计算平均值代表灯丝电阻无意义,小刘忽略温度对电阻的影响.
L1换成定值电阻,能保持电阻不变,通过改变电压来改变电流能探究电流跟电压的关系.
故答案为:(1)如上图;(2)左,电流(A );(3)大;(4)电压;(5)温度;正确.

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