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When a capacitor is charged after 1 time constant the v across the capacitor is?
Wiki User
∙ 13y ago
Depending on the circuit, 63% of the available voltage.
Wiki User
∙ 13y ago
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What would happen to the current flow in a dc circuit when the capacitor is fully charged?
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/
A 10 microfard capacitor is charged through a 10 ohm resistor. How long will it take for the capacitor t fully charge?
If a 10 microfarad capacitor is charged through a 10 ohm resistor, it will theoretically never reach full charge. Practically, however, it can be considered fully charged after 5 time constants. One time constant is farads times ohms, so the time constant for a 10 microfarad capacitor and a 10 ohm resistor is 100 microseconds. Full charge will be about 500 microseconds.
What rule of thumb is used to indicate the amount of time it will take a capacitor to charge to the source voltage if there is a resistor in series with the capacitor?
A: Mathematically speaking the capacitor will never charge to the source because it takes one time constant to reach 63% and so on but for practical uses it is assume to be fully charged in 5 time constants R X C = 1 TIME CONSTANT
How do you change time constant of capacitor?
In theory ... on paper where you have ideal components ... a capacitor all by itself doesn't have a time constant. It charges instantly. It only charges exponentially according to a time constant when it's in series with a resistor, and the time constant is (RC). Keeping the same capacitor, you change the time constant by changing the value of the resistor.
Will voltage drop across a capacitor?
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.
What would happen to the current flow in a dc circuit when the capacitor is fully charged?
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/
How do you figure out the charge of a capacitor?
A: from a voltage source a capacitor will charge to 63 % of the voltage in one time constant which is define the voltage source Resistance from the source time capacitor in farads. it will continue to charge at this rate indefinitely however for practical usage 5 time constant is assume to be fully charged
A 10 microfard capacitor is charged through a 10 ohm resistor. How long will it take for the capacitor t fully charge?
If a 10 microfarad capacitor is charged through a 10 ohm resistor, it will theoretically never reach full charge. Practically, however, it can be considered fully charged after 5 time constants. One time constant is farads times ohms, so the time constant for a 10 microfarad capacitor and a 10 ohm resistor is 100 microseconds. Full charge will be about 500 microseconds.
What rule of thumb is used to indicate the amount of time it will take a capacitor to charge to the source voltage if there is a resistor in series with the capacitor?
A: Mathematically speaking the capacitor will never charge to the source because it takes one time constant to reach 63% and so on but for practical uses it is assume to be fully charged in 5 time constants R X C = 1 TIME CONSTANT
How do you change time constant of capacitor?
In theory ... on paper where you have ideal components ... a capacitor all by itself doesn't have a time constant. It charges instantly. It only charges exponentially according to a time constant when it's in series with a resistor, and the time constant is (RC). Keeping the same capacitor, you change the time constant by changing the value of the resistor.
What happens to the current in a circuit as a capacitor charges?
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.
Will voltage drop across a capacitor?
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.
IF capacitor that has been connected across a battery for comparatively long time becomes?
If the capacitor isn't punctured or failed, then it becomes charged to the voltage of the battery almost immediately after it's connected to it, and stays that way.
What is function of capacitor in 555 IC?
the capacitor and its associated resistor set the time constant.
How long does it take to be fully charged when a 500uF capacitor is in series with a 2.7 Kohm resistor?
Equation for voltage across capacitor in series RC circuit is as follow, vc = V(1-e-t/RC) V = DC voltage source. So theoretically time taken for capacitor to charge up to V volt is INFINITY. But practically we assume 95% or 98% of source voltage as fully charge. RC is the time constant which is the time take for capacitor to charge 63%. In this case time constant is 500uF*2.7Kohm = 1.3sec Time taken to charge 95% = 3*T = 3*1.3 = 3.9sec T = time constant Time taken to charge 98% = 4*T = 4*1.3 = 5.2sec
Why is capacitor leakage a problem in a RC timer circuit?
Because the timing is set by the time constant of a resistor and a capacitor. With R in ohms and C in Farads, the time-constant is RC in seconds. If the capacitor leaks the timing will be wrong.
What is the act of recovering energy from Capacitor?
A: It is called discharging a capacitor. The charge will follow the rules of a time constant set up by the series resistor and the capacitor. 1 time constant 63% of the charge will be reached and continue at that rate.
