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How much time does a hyperactive RC charge?

Updated: 8/20/2019
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12y ago

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30-50 min

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12y ago
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Q: How much time does a hyperactive RC charge?
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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


Time constant of an RC circuit increases if the value of the resistance is?

Answer : increase The time required to charge a capacitor to 63 percent (actually 63.2 percent) of full charge or to discharge it to 37 percent (actually 36.8 percent) of its initial voltage is known as the TIME CONSTANT (TC) of the circuit. Figure 3-11. - RC time constant. The value of the time constant in seconds is equal to the product of the circuit resistance in ohms and the circuit capacitance in farads. The value of one time constant is expressed mathematically as t = RC.


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-- The quantity 'RC' has the physical dimensions of Time. -- If the capacitor is charging through a resistor, then 'RC' is the time it takes to charge up to (1 - 1/e) of the voltage it still has to go to become fully-charged. -- If the capacitor is discharging through a resistor, then 'RC' is the time it takes to discharge to 1/e of its present voltage. -- ' e ' is the base of natural logarithms, approximately 2.71828... -- 'RC' is called the 'time constant' of the resistor/capacitor combination.


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The time constant for a RC circuit is given by?

In an RC circuit the time constant is found by R x C. T = R x C to be precise.It is the time required to charge the capacitor through the resistor, to 63.2 (≈ 63) percent of full charge; or to discharge it to 36.8 (≈ 37) percent of its initial voltage. These values are derived from the mathematical constant e, specifically 1 − e − 1 and e − 1 respectively.


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A: Assuming RC filter the capacitor will charge trough the resistor for time = r x c at the first pulse the capacitor is zero and charges to 63% or tries to if the pulse has at least 5 times the RC time if not it will try to charge to 63% and discharge at the same rate therefore reaching an equilibrium if the pulses are symmetrical


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What happens to the time taken for the capacitor to charge and discharge when you increase the resistance in the circuit?

It increases. The time constant of a simple RC circuit is RC, resistance times capacitance. That is the length of time it will take for the capacitor voltage to reach about 63% of a delta step change. Ratio-metrically, if you double the resistance, you will double the charge or discharge time.


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