问题 问答题

如图所示,电路中R0为定值电阻,R为滑动变阻器,总阻值为R,当在电路两端加上恒定电压U,移动R的滑片,可以改变电流表的读数范围为多少?

答案

设滑动变阻器滑动触头左边部分的电阻为Rx.电路连接为R0与Rx并联,再与滑动变阻器右边部分的电阻R-Rx串联,

干路中的电流:I=

U
R+R-RX
=
U
R0RX
R0+RX
+R-RX

电流表示数:I′=

R0
R0+RX
I=
R0
R0+RX
U
R0RX
R0+RX
+R-RX
=
UR0
RR0+RRX-
R2X
=
UR0
RR0+(
R
2
)2-(RX-
R
2
)2

由上式可知:当Rx=

R
2
时,I最小为:Imin=
4UR0
4RR0+R2
;当Rx=R或Rx=0时,I有最大值,Imax=
U
R
; 

即电流表示数变化范围为:

4UR0
4RR0+R2
U
R

答:电流表示数变化范围是::

4UR0
4RR0+R2
U
R

单项选择题
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