试题与答案

如图所示的电路,将AB两端接入10V电源,电压表示数为3V,拆去AB两端电源,再

题型:选择题

题目:

如图所示的电路,将AB两端接入10V电源,电压表示数为3V,拆去AB两端电源,再将CD两端接入10V电源,电压表示数为7V,则R1:R3的值为(  )

A.1:1

B.7:3

C.49:9

D.100:21

答案:

答案:C

题目分析:(1)将AB两端接入10V电源时,R1与R2串联,电压表测R2两端的电压,根据串联电路的电压特点求出R1两端的电压,利用串联电路的电流特点和欧姆定律求出R1与R2的比值;

(2)拆去AB两端电源,再将CD两端接入10V电源,R2与R3串联,电压表测R2两端的电压,根据串联电路的电压特点求出R3两端的电压,利用串联电路的电流特点和欧姆定律求出R2与R3的比值,进一步求出R1:R3的值.

(1)将AB两端接入10V电源时,R1与R2串联,电压表测R2两端的电压,

∵串联电路中总电压等于各分电压之和,

∴R1两端的电压U1=U﹣U2=10V﹣3V=7V,

由欧姆定律可得:

====,

拆去AB两端电源,再将CD两端接入10V电源,R2与R3串联,电压表测R2两端的电压,

R3两端的电压U3=U﹣U2′=10V﹣7V=3V,

====

=×=×=

点评:本题考查了串联电路的特点和欧姆定律的应用,分清电源不同位置时的连接方式和电压表测量的电路元件是关键.

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