capacitance
DC,
AC, and
AC are obtained through a rectifier. The
voltage changes with time.
Why do you need to use capacitors to
maintain the working voltage? The voltage is above a fixed value during operation. If the voltage is lower than the fixed value for a certain period of time, the electric appliance will turn off, because the voltage after passing through the rectifier is sometimes less than A fixed value, using a capacitor to achieve
how large the current is at the capacitor where the electrical appliance does not go out .
How to
figure out the relationship between the capacitor and the voltage? Know that
i = △q/ △T The change of charge per unit time/unit time
q = Capacitance * Voltage here only Knowing that the capacitor voltage is proportional
to the current ,
i = △ (C * U) / △ T = △ C / △ T * U + △ U / △ T * C The
capacitance will not change over time and the capacitance is a fixed value, so △ C For 0
, i=△ U / △ T * C
The formula
i = c * △U / △T Current = Capacitance * The voltage that changes per unit time/
How much current does the capacitor short-circuit produce
when the capacitor is short-circuited momentarily
The voltage will gradually decrease with time. When the time is 0, k is the voltage.
When the voltage is 0, the voltage/time with a negative slope m
takes Uc = -U/t into ic=c* △Uc /△t When the capacitor is short-circuited, i =-c * △U / △T.
Since the short-circuit occurs for a short time, the current will be infinite if it is ideal.
How big
is the voltage at the capacitor The picture shows the voltage change at the capacitor. At the beginning, the voltage at the capacitor is a fixed value, and then a fixed-size voltage source is added. At this time, the voltage situation is as shown in the figure below.
Formula
Voltage at the capacitor = voltage at the beginning + t1 to t2 Sum/capacitance of current over time