初三物理电功率题 要详解 要快 在线等 好的追加分
图所示的电路中,电源电压保持不变,R1=R3=12Ω,当S1S2都断开时,电压表的示数为U1,电流表的示数是I,电阻R2消耗的功率是3w,当S1闭合,S2断开时,电流表的...
图所示的电路中,电源电压保持不变,R1=R3=12Ω,当S1S2都断开时,电压表的示数为U1,电流表的示数是I,电阻R2消耗的功率是3w,当S1闭合,S2断开时,电流表的示数是3I,电压表的读数是U2,求;
(1)U1:U2;
(2)S1S2都闭合时,电流表的示数是? 展开
(1)U1:U2;
(2)S1S2都闭合时,电流表的示数是? 展开
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分析:当S1S2都断开时,串联。电路总电阻R=R1+R2=12+R2,总电流为I。
当S1闭合,S2断开时,混联(有串联有并联)总电阻R'=R1R3/(R1+R3)+R2=6+R2,总电流3I。
由于电源电压保持不变,可得U电=RI=3IR' ············12+R2=3(6+R2),R2=-3
额··················································································································································································
当S1S2都断开时,这时是一个R1与R2的串联电路,电流处处相等,电压与电阻成正比。
P(R2)=U(R2)I
当S1闭合,S2断开时,混联(有串联有并联)总电阻R'=R1R3/(R1+R3)+R2=6+R2,总电流3I。
由于电源电压保持不变,可得U电=RI=3IR' ············12+R2=3(6+R2),R2=-3
额··················································································································································································
当S1S2都断开时,这时是一个R1与R2的串联电路,电流处处相等,电压与电阻成正比。
P(R2)=U(R2)I
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