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basically a capacitor will charge to the input DC level however it will mathematically never happen since capacitors charge at a certain rate the voltage drop across a capacitor will follow the R C time constant or 63% of the applied voltage for a unit time.AnswerIn the case of an a.c. supply, yes, there will be a voltage drop across a capacitor. In the case of an 'ideal' capacitor, this will be the product of the load current and the capacitive reactance of the capacitor.
The amperage of the circuit increases and the voltage drop across the appliances will tend to increase.
You can always use a higher voltage rated capacitor, it will probably just last a little longer.
A capacitor charge as a time constant of R resistance C capacitance in ufd and it is defined as 63% for one time constant for the constant voltage source. Electronic engineers assume that a capacitor is fully charged by a 5 times constant. however mathematically speaking it will never be fully charged for obvious reasons. Therefore the answer is current will never stop/
It will rotate faster.Since the fan is not designed for this, it will progressively ruin it.
series resonant circuit
Most likely not. The duration and current supplied by a capacitor in the microFarad range would be short and small, respectively. On the other hand, of one were touched by the capacitor leads on the eyeball, really thin-skinned areas, or for some reason across the chest over the heart, some damage and/or great pain could happen.
basically a capacitor will charge to the input DC level however it will mathematically never happen since capacitors charge at a certain rate the voltage drop across a capacitor will follow the R C time constant or 63% of the applied voltage for a unit time.AnswerIn the case of an a.c. supply, yes, there will be a voltage drop across a capacitor. In the case of an 'ideal' capacitor, this will be the product of the load current and the capacitive reactance of the capacitor.
Nothing will happen, you can go over 6.3v but never under......
If the capacitor is charged then the battery will explode.
It flows out of the capacitor into the external circuit
It will increase the ripple factor that the capacitor is in the circuit to smooth out.
the diode when forward biased will conduct and during reverse biased condition(generally doring reverse biased condition ckt is open mens no current flows;when register is connected)current flows but during reverse biased condition 1--for sometimes initially current flows due to discharging of capacitor. 2--then ckt will be having no current
Oversizing the capacitor can shorten motor life.
When the terminals of a capacitor are connected together, the capacitor will discharge, returning to a zero potential state. Capacitors resist voltage change, meaning that if the capacitor is in a circuit that has zero voltage potential, the capacitor will eventually achieve zero potential. If the capacitor is in a circuit that has a 5 volt potential, the capacitor will seek and attempt to maintain that 5 volt potential (provided that the capacitor is rated at 5 volts or more). In an AC circuit, the capacitor will tend to smooth out the sin wave of the current, resisting change in both directions. In a DC power supply circuit, a capacitor will tend to reduce the voltage "ripple", and if the circuit is designed properly, will provide a smooth DC voltage. Shorting the terminals of a capacitor is effectively what often happens in many circuits; it's not a problem.
I think you over-laod and blow the capacitor
The amperage of the circuit increases and the voltage drop across the appliances will tend to increase.