怎样求滑动变阻器接入电路中的最大阻值及阻值范围

2025-04-05 19:54:38
推荐回答(1个)
回答1:

串联电路中,当滑动变阻器的阻值等于与其串联的电阻阻值时,滑动变阻器消耗的电功率最大.
即滑动变阻器与定值电阻平分电压时,滑动变阻器消耗功率的功率为最大值,其最大值为U²/4R1.(R1为定值电阻的阻值)
补充:
证明:
令滑动变阻器接入电路电阻的功率Px为因变量,滑动变阻器接入电路电阻Rx为自变量,则
串联电路的总电阻:R总=R1+R2
滑动变阻器接入电路电阻的两端电压为:U2=R2U/(R2+R1)
滑动变阻器接入电路电阻的功率:P2=U2²/R2=[R2²U²/(R2+R1)²] × (1/R2)
即:P2=[R2/(R2+R1)²] U²=[1/(R2+2R1+R²1/R2)] U² .(1)
令Y=R2+2R1+R1²/R2,设R2=x,R1=a,则原式可变为:
y=x+2a+a²/x
=2a+(√x)²+(a/√x)²-2(√x)(a/√x)+2(√x)(a/√x)
=2a+[√x-(a/√x)]²+2(√x)(a/√x)
=4a+[√x-(a/√x)]²
当√x-(a/√x)=0时,上式取得最小值,即当x=a时,y取得最小值,其最小值为4a.
根据上面所设的关系,即当R2=R1时,Y取得最小值,即最小值Y=4R1
而当Y取得最小值时,1/(R2+2R1+R1/R2) 即可取得最大值,则分析上面(1)式可知,此时P2取得最大值,P2的最大值为:P2=U²/4R1
综上所述,当R2=R1时,即滑动变阻器与定值电阻平分电压时,滑动变阻器消耗功率的功率为最大值,其最大值为U²/4R1.

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