电容元件特性电压连续性★Vc(t):lo+AtJi(t)dt = 0= Vc(folim(t.CAt->0-41n当电容电流（功率）为有限值时，电容电压不能发生突变21CireuitAnalvsis by Benjing Jiaotong Uiniversity
Circuit Analysis by Beijing Jiaotong University 电容元件特性 21 电压连续性✮ 当电容电流（功率）为有限值时，电容电压不能发生突变。 ( ) ( ) + − 0 = 0 v t v t C C ( ) + − → = t t t t c t i d C 0 0 0 1 0 lim ( ) − = t i d C vC t C 1 ( )
电容元件例题例求电容上的电压。i(t) (mA)v(0)=()dtCC =5μF20Vv(0)=0Vt<0t (ms)-1023450≤t<2ms(a)v(t)(V)7(t)20×10-3dt +v(0)5×10-68v()= 4000tt (ms)2-10413t≥2ms v(t)=8V(b)22北京交通大学国家电工电子教学基地电路理论系列课程组
北京交通大学 国家电工电子教学基地 电路理论系列课程组 22 求电容上的电压。 ( ) ( ) − = t i d C v t 1 v t( ) = 0V ( ) + = − − t v t d v 0 3 6 20 10 (0) 5 10 1 0 2ms t v(t) = 4000t t 2ms v t( ) = 8V t 0 i C C = 5 mF v 例 电容元件例题
电容元件例题C =1FdvHHOOVi(t)v(0)=i(t)dtdt1+V-=(c)+i(t)d?动态特性记忆特性v/VLilA2t/s初值t/s08205642i(t)dtv(to) +v(tv/Vto,v(to)NilA=v(to)t/st/s?780561234567d电容电压连续性tofot25sisbyBellinsLUiniversityr
Circuit Analysis by Beijing Jiaotong University 25 i v C =1F v / V t / s −12 1 0 1 2 3 4 5 6 − 2 7 8 i / A t / s −11 0 1 2 4 5 6 2 3 7 8 动态特性 电容元件例题 i / A t / s −11 0 1 2 4 5 6 2 3 7 8 v / V t / s −12 1 0 1 2 3 4 5 6 − 2 7 8 ( ) ( ) 0 0 1 = + d t v t i t C ( ( )) 0 0 t , v t 初值 记忆特性 ( ) = dv i t C dt ( ) ( ) 1 − = t v t i d C − 电容电压连续性 0 t + 0 0 t t ( ) ( )d 1 ( ) ( ) 0 0 0 0 0 − + − = = + + − v t i C v t v t t t
电容元件例题C =1F储能特性（能量）dvOCi(t)dtidy+Vo(t)dt:ddt储能特性（功率）-CyC?(t)-8V/V2dyt/sp=vi=CyCy?(t)078dtA16122Ap/WAW/JilA422t/st/ st/s01.2h0842356.726CircuitAunalvsis by Beuing Jiaotong Uiniversity
Circuit Analysis by Beijing Jiaotong University 26 d d = = v p vi Cv t v / V t / s −12 1 0 1 2 3 4 5 6 − 2 7 8 i / A t / s −11 0 1 2 4 5 6 2 3 7 8 p / W 2 t / s 0 1 2 3 5 6 − 24 4 7 8 储能特性（功率） 电容元件例题 储能特性（能量） 2 2 2 d ( )d d d 1 1 ( ) ( ) 2 2 1 ( ) 2 − − = = = − − = t t v w p v Cv t Cv Cv t w/J 1 t / s 0 1 2 3 5 6 2 4 7 8 i v C =1F ( ) = dv i t C dt
电容的串并联等效并联：i=i+iC= C, +C2dv北=(C)dt+v-串联：111C,CC,V= V,1+ViHO+i(t)dt+V-+V2-+1XV27CircuitAunalvsis bv Beuing Jiaotong University
Circuit Analysis by Beijing Jiaotong University 电容的串并联等效 27 t v C C i i i d d ( ) 1 2 1 2 = + = + 1 i C1 2 i C2 v i i v C = C1 +C2 i v 1 2 1 1 1 C C C = + − = + = + t i C C v v v ) ( )d 1 1 ( 1 2 1 2 C1 C2 i 1 v 2 v v 并联： 串联：