Time constant in an RC filter is resistance times capacitance. With ideal components, if the resistance is zero, then the time constant is zero, not mattter what the capacitance is. In a practical circuit, there is always some resistance in the conductors and in the capacitor so, if the resistance is (close to) zero, the time constant will be (close to) zero.
A capacitor will appear to be an open circuit to a DC source, but only after equilibrium is reached. Proof: A capacitor resists a change in voltage. The equation is ... dv/dt = i/c ... which means that the rate of change of voltage in volts per second is equal to current in amperes divided by capacitance in farads. A DC source has constant voltage. If you charge a capacitor to a constant voltage, then, at equilibrium, dv/dt is zero. This means that i must also be zero, since c is not zero. Ohm's law states that resistance is voltage divided by current. The limit of this is that, when current is zero, then resistance must be infinity. Therefore, the capacitor will have infinite resistance and appear to be open circuited.
Consider the instantaneous DC analysis. Initially, the capacitor has zero resistance. You apply a voltage and current is controlled by other resistive elements alone. As the capacitor charges, its effective resistance rises. This adds to the net resistance in the circuit, reducing current. At full charge, the capacitor has infinite resistance, so there is no current. Remember that the equation for a capacitor is dv/dt = i/c.
Charge the capacitor. Potential difference is a scientific term for what is more commonly called voltage. ANSWER: If big enough the battery will see a short initially and then proceed to charge the capacitor at a rate of 63% of the voltage in one time constant defined as RC For engineering purposes after 5 time the time constant the battery will and the capacitor zero potential different. The proper term should be virtual no difference.
We can guess that the resistor is used for discharging the capacitor's plates. Generally we short the two terminals on a capacitor to discharge it fully. A resistor will take more time to do this than shorting-out the terminals: the higher the resistance, the longer the time that will be taken to discharge a capacitor fully.
because resistance is restricting the current and voltage, so for it be accurate you need to know what the voltage and the amps are.AnswerCapacitance is quite independent of resistance and, therefore, it will NOT vary if resistance is changed.
A capacitor will appear to be an open circuit to a DC source, but only after equilibrium is reached. Proof: A capacitor resists a change in voltage. The equation is ... dv/dt = i/c ... which means that the rate of change of voltage in volts per second is equal to current in amperes divided by capacitance in farads. A DC source has constant voltage. If you charge a capacitor to a constant voltage, then, at equilibrium, dv/dt is zero. This means that i must also be zero, since c is not zero. Ohm's law states that resistance is voltage divided by current. The limit of this is that, when current is zero, then resistance must be infinity. Therefore, the capacitor will have infinite resistance and appear to be open circuited.
When the frequency is zero(i.e when dc power is supplied), capacitor is open is treated as open circuit having infinite resistance.
Consider the instantaneous DC analysis. Initially, the capacitor has zero resistance. You apply a voltage and current is controlled by other resistive elements alone. As the capacitor charges, its effective resistance rises. This adds to the net resistance in the circuit, reducing current. At full charge, the capacitor has infinite resistance, so there is no current. Remember that the equation for a capacitor is dv/dt = i/c.
Charge the capacitor. Potential difference is a scientific term for what is more commonly called voltage. ANSWER: If big enough the battery will see a short initially and then proceed to charge the capacitor at a rate of 63% of the voltage in one time constant defined as RC For engineering purposes after 5 time the time constant the battery will and the capacitor zero potential different. The proper term should be virtual no difference.
The charging time of a capacitor is usually lower than the discharging time because during charging, the voltage across the capacitor is increasing from zero to its maximum value, which initially allows a higher current to flow. During discharging, the voltage across the capacitor is decreasing from its maximum value to zero, resulting in a lower current flow. This difference in current flow affects the time it takes for the capacitor to charge and discharge.
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.
We can guess that the resistor is used for discharging the capacitor's plates. Generally we short the two terminals on a capacitor to discharge it fully. A resistor will take more time to do this than shorting-out the terminals: the higher the resistance, the longer the time that will be taken to discharge a capacitor fully.
because resistance is restricting the current and voltage, so for it be accurate you need to know what the voltage and the amps are.AnswerCapacitance is quite independent of resistance and, therefore, it will NOT vary if resistance is changed.
If traveling at constant speed in a constant direction then net force is zero as there is no acceleration. Acceleration would change one or the other, or both. F = ma = m (0) = 0
Yes, but the net force is ZERO! If an object is moving at constant velocity, the sum of the forces acting upon it is zero. If at any time the sum of the forces -- sometimes called the net force -- is non-zero, the object will accelerate in the direction of the resultant force.
Depending on the value, it varies. Using a multimeter set for Kilohms resistance, check that it appears open circuit. This may be hard to determine, due to other components in circuit. It should NOT show zero resistance. Electrolytic capacitors above 1 microFarad will show some storage of charge. Using the meter with the probes one way, there should be a percepible rise of resistance. Reverve the leads and you should get an intial negative value followed by a slow rise in resistance. If there is no perceptible rise, it could be open circuit. If it shows Zero, it will indicate a dead short.
because in a capacitor only charges are stored so the stored charges are gives the zero current