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.
It's not.
The voltage drop across an LED varies, but is typically around 2 volts.
Depending on the circuit, 63% of the available voltage.
(a) what is the total capacitance of this arrangement (B) the charge stored on each capacitor (C) the voltage across the 50 micro farad capacitor and the energy stored in it. 20v and 20+30+50 micro farad
Voltage across all parallel capacitor's is same i.e. it is equal to supply voltage, it can be measured using digital volt meter (any high input impedance volt meter). When capacitors are in series; voltage drop depends on charge stored in the capacitor. it can be given by the formula V x V = 2 / (joules x capacitance). This voltage can also be measured using digital volt meter.
Because the capacitor discharges. so voltage across the capacitor decreases.
What happens to the current in a circuit as a capacitor charges depends on the circuit. As a capacitor charges, the voltage drop across it increases. In a typical circuit with a constant voltage source and a resistor charging the capacitor, then the current in the circuit will decrease logarithmically over time as the capacitor charges, with the end result that the current is zero, and the voltage across the capacitor is the same as the voltage source.
If a bypass capacitor is used the voltage drop across emitter resistance is reduced which in turn increases the gain.....
In a capacitor, the current LEADS the voltage by 90 degrees, or to put it the other way, the voltage LAGS the current by 90 degrees. This is because the current in a capacitor depends on the RATE OF CHANGE in voltage across it, and the greatest rate of change is when the voltage is passing through zero (the sine-wave is at its steepest). So current will peak when the voltage is zero, and will be zero when the rate of change of voltage is zero - at the peak of the voltage waveform, when the waveform has stopped rising, and is about to start falling towards zero.
You charge a capacitor by placing DC voltage across its terminal leads. Make sure when using a polarized capacitor to place positive voltage across the positive lead (the longer lead) and negative voltage across the negative lead. Also make sure that the voltage you charge the capacitor to doesn't exceeds its voltage rating.
In order to double the voltage across a capacitor, you need to stuff twice as much charge into it.
fully charged.
the voltage number on the capacitor indicates that the capacitor can with stand to that particular voltage across it.generally during design, the value of capacitor will be selected in such a way that this voltage rating should be double than what really we get in the circuit
It's not.
The voltage drop across an LED varies, but is typically around 2 volts.
Lets have an example of simple RC high pass filter. Here, we take output across Resistor(in HPF). The tilt is because of charging of capacitor. you can say, as capacitor charges (ofcourse with voltage) the same amount of voltage has to drop across resistor ( to follow KVL). Since we are taking output across Resistor, so we see small voltage drop (as tilt). This can be minimized by keeping RC time constant large.
Eventually, the capacitor will charge to approximately the source voltage level. As this occurs, the current in the circuit will drop to near zero.