在“测定小灯泡的电阻”实验中,有如下器材:电压表、电流表、开关、电压为6V的电源、正常发光时电压是3.

2025-04-03 06:03:49
推荐回答(1个)
回答1:

(1)灯泡正常发光时,电流为I=

U
R
=
3.8V
10Ω
=0.38A,所以电流表可选择0~0.6A量程,即将电流表0.6A接线柱与灯泡的左接线柱相连,将开关的左接线柱与滑动变阻器下面任一接线柱相连,如图所示:

(2)连接电路时,为了安全,开关必须断开,滑动变阻器的滑片移动阻值最大处,即滑片放在最右端;
(3)为保护电路,连接实物图时应:①断开开关,②滑动变阻器的滑片要调到使连入电路的电阻为最大的位置. 甲组同学连接好最后一根导线,灯泡立即发出明亮耀眼的光并很快熄灭,说明连完最后一根线,电路立即接通且电路电流较大,电路阻值太小,原因是:①连接电路时没断开开关;②滑动变阻器没调到最大阻值.
(4)A、灯泡断路时,灯泡不亮,电流表无示数,不符合题意;
B、灯泡短路时,灯泡不亮,电流表有示数,符合题意;
C、滑动变阻器短路时,灯泡亮,电流表有示数,不符合题意;
D、滑动变阻器连入的电阻太大时,电路中电流太小,灯泡的实际功率太小,因此灯泡不亮,电流表有示数,符合题意.
故选BD.
(5)根据表中数据可知三次小灯泡的阻值不相等,因为灯泡两端电压不同时,灯丝的温度不同,因此灯丝的电阻不同,即灯丝电阻与温度有关.
(6)电流表量程为0~0.6A,分度值为0.02A,故示数为0.2A;
电压表量程为0~3V,分度值为0.1V,故示数为2V;
由I=
U
R
可知,被测电阻的阻值为:R=
U
I
=
2V
0.2A
=10Ω.
故答案为:(2)断开;右;(3)①连接电路时没断开开关;②滑动变阻器没调到最大阻值;(4)BD;(5)不同;灯丝电阻与温度有关;(6)0.2;2;10.

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