电容器C 1=6微法,C2=3微法,电阻R1=6欧,R2=3欧.求(1)当开关断开时AB间电压,(电容器C 1=6微法,C2=3微法,电阻R1=6欧,R2=3欧.求(1)当开关断开时AB间电压,(2)当S闭合时,电容器C1的电荷量改变量.
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![电容器C 1=6微法,C2=3微法,电阻R1=6欧,R2=3欧.求(1)当开关断开时AB间电压,(电容器C 1=6微法,C2=3微法,电阻R1=6欧,R2=3欧.求(1)当开关断开时AB间电压,(2)当S闭合时,电容器C1的电荷量改变量.](/uploads/image/z/3485583-63-3.jpg?t=%E7%94%B5%E5%AE%B9%E5%99%A8C+1%3D6%E5%BE%AE%E6%B3%95%2CC2%3D3%E5%BE%AE%E6%B3%95%2C%E7%94%B5%E9%98%BBR1%3D6%E6%AC%A7%2CR2%3D3%E6%AC%A7.%E6%B1%82%281%29%E5%BD%93%E5%BC%80%E5%85%B3%E6%96%AD%E5%BC%80%E6%97%B6AB%E9%97%B4%E7%94%B5%E5%8E%8B%2C%28%E7%94%B5%E5%AE%B9%E5%99%A8C+1%3D6%E5%BE%AE%E6%B3%95%2CC2%3D3%E5%BE%AE%E6%B3%95%2C%E7%94%B5%E9%98%BBR1%3D6%E6%AC%A7%2CR2%3D3%E6%AC%A7.%E6%B1%82%281%29%E5%BD%93%E5%BC%80%E5%85%B3%E6%96%AD%E5%BC%80%E6%97%B6AB%E9%97%B4%E7%94%B5%E5%8E%8B%2C%282%29%E5%BD%93S%E9%97%AD%E5%90%88%E6%97%B6%2C%E7%94%B5%E5%AE%B9%E5%99%A8C1%E7%9A%84%E7%94%B5%E8%8D%B7%E9%87%8F%E6%94%B9%E5%8F%98%E9%87%8F.)
电容器C 1=6微法,C2=3微法,电阻R1=6欧,R2=3欧.求(1)当开关断开时AB间电压,(电容器C 1=6微法,C2=3微法,电阻R1=6欧,R2=3欧.求(1)当开关断开时AB间电压,(2)当S闭合时,电容器C1的电荷量改变量.
电容器C 1=6微法,C2=3微法,电阻R1=6欧,R2=3欧.求(1)当开关断开时AB间电压,(
电容器C 1=6微法,C2=3微法,电阻R1=6欧,R2=3欧.求(1)当开关断开时AB间电压,(2)当S闭合时,电容器C1的电荷量改变量.
电容器C 1=6微法,C2=3微法,电阻R1=6欧,R2=3欧.求(1)当开关断开时AB间电压,(电容器C 1=6微法,C2=3微法,电阻R1=6欧,R2=3欧.求(1)当开关断开时AB间电压,(2)当S闭合时,电容器C1的电荷量改变量.
当开关断开时AB间电压为电源电压U,理由电阻R1、R2 I=0故无压降
C1电荷量为U*6微法(库仓)
当S闭合时 C1与R2并联 UC1电压为U*R2/(R1+R2),故当S闭合时C1电荷量为UC1*6微法(库仓)
即二者相减即为C1的电荷量改变量
因图不清以上半部为C1R1、下半部为C2R2、
数字计算结果简化因较简单我就不做了,
(1)当S断开时,电路稳定时,电路中没有电流,φB=φC,φA=φD,UAB=UCD=U.
(2)S断开时,UC1=U,Q1=C1Uc1=C1U
S闭合时,UC1=UCB=UR1/(R1+R2)=2U/3
Q1'=2C1U/3
△Q=Q1-Q1'=C1U-2C1U/3=C1U/3麻烦您分析一下S 闭合或断开,电子怎么移动。谢谢U的正负极不知,没办法判断假设一边为负极...
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(1)当S断开时,电路稳定时,电路中没有电流,φB=φC,φA=φD,UAB=UCD=U.
(2)S断开时,UC1=U,Q1=C1Uc1=C1U
S闭合时,UC1=UCB=UR1/(R1+R2)=2U/3
Q1'=2C1U/3
△Q=Q1-Q1'=C1U-2C1U/3=C1U/3
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