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1、欧姆定律单元测试题三(参考答案)DIV.MyFav_1299808169694 P.MsoNormalTEXT-JUSTIFY: inter-ideograph; FONT-SIZE: 10.5pt; MARGIN: 0cm 0cm 0pt; COLOR: windowtext; FONT-FAMILY: Times New Roman; TEXT-ALIGN: justifyDIV.MyFav_1299808169694 LI.MsoNormalTEXT-JUSTIFY: inter-ideograph; FONT-SIZE: 10.5pt; MARGIN: 0cm 0cm 0pt; COL
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9、v_1299808169694 DIV.Section1page: Section1一、130;03;30;0;30 26;2;1:2;1:1 3变大;不变;不变;45;075;5S1、S2;断开;655;1100;220;7导体;1200二、1B;2C;3B;4B;5D;6C;7C;8A;9C;9C三、1010;018;030042;050;(2)51 (3)调节导体两端电压和通过导体的电流。2电阻一定时,导体中的电流与导体两端的电压成正比;电压一定时,导体中的电流与导体的电阻成反比。3电流表改用06A的量程;可以使测量结果更准确。4(1)略;(2)024;75;无法得到要测量的电阻的阻值;(3)至少应移动变阻器滑片再测两组电压和电流值。进行多次测量求平均值使测量结果更准确。五、1(1)U1=I1R1=1A5=5V;(2)I2=I-I1=15A-1A=05A;(3)U2=U1=5V R2=U2/I2=5V /05A=52电源电压高于灯泡正常工作,应采用串联分压的原理进行分压,因此电路中应串联一个电阻R。灯泡两端电压U1为25V,则串联的电阻R两端的按压U2=U-U1=55V-25V=3V 则根据分压原理U1/R1=U2/R2=25V/10=3V/R2R2=12 所以,应在电路中串联一个阻值为12的电阻。附加题:1(1)133;20;24(2)小灯泡的电阻随温度的升高而增大。2略