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设C_1、C_2两个电容器串联接在电压为U的恒压源上,求C_1、C_2两端的电压U_1U_2。教科书上给出的答案是: U_1=C_2/(C_1+C_2)U (1) U_2=C_1/(C_1+C_2)U (2)但实际情况是: U_1=R_1/(R_1+R_2)U (3) U_2=R_2/(R_1+R_2)U (4)这里R_1、R_2是C_1、C_2的漏电电阻。实用电容器的漏电电阻不是无穷大,而是有限值,而且随介质的不同可以相差好几个数量级。例如高压云母电容器的漏电电阻大于10~(12)欧,而低压纸质电容器则常小于10~9欧。
Let C_1, C_2 two capacitors connected in series on the voltage source U of the constant voltage, seeking C_1, C_2 voltage U_1U_2. The answer given in the textbook is: U_1=C_2/(C_1+C_2)U (1) U_2=C_1/(C_1+C_2)U (2) But the actual situation is: U_1=R_1/(R_1+R_2)U ( 3) U_2=R_2/(R_1+R_2)U (4) R_1 and R_2 are the leakage resistances of C_1 and C_2. The leakage resistance of a practical capacitor is not infinity, but a finite value, and can vary by several orders of magnitude depending on the medium. For example, leakage resistance of high voltage mica capacitors is greater than 10~(12) ohms, while low voltage paper capacitors are often less than 10~9 ohms.