(1) E^2/(6+1.5+r) = 2
E^2/(3//6+1.5+r) = 4
解出 电源的电动势 E = 4V
内电阻 r = 0.5 Ω;
(2)闭合S时,电源的输出功率为
(4/(3//6+1.5+0.5))^2 * (3//6+1.5) = 3.5 W
(3) S断开时,电容器两端电压为 U = 4*6/(6+1.5+0.5) = 3 V
所带的电荷量为 Q = CU = 20*10^(-6)*3 = 6*10^(-5) C
S闭合时,电容器两端电压为 U = 0
所带的电荷量为 Q = CU = 0
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