I have studied the fact that the capacitor does not like a varying voltage across its terminals, and it acts as a short circuit initially so that there is no voltage drop across it. If this is true, then why does it acts a short circuit (or a low impedance connection) in case of AC signals, which have alternating or changing voltages every moment?
Electronic – High frequency response of capacitors
accapacitordcfrequency response
Related Solutions
The first circuit in your question only exists in badly designed homework problems.
The reason is that an ideal capacitor in parallel with an ideal voltage source does nothing. The ideal voltage source can supply whatever current is needed to drive the capacitor to it's level without sagging, so the capacitor provides no filtering.
but...
In the real world there are no ideal voltage sources. They have series resistance. Or if they're located far from the point of load, there's a series inductance between the source and the load. This is the scenario where a parallel capacitor is useful.
Combined with the voltage source's parasitic resistance, the parallel capacitor provides filtering and reduces noise at the load.
Also, for more complex loads than the simple resistor in your circuit, if the load current varies (for example if it's a digital logic chip with it's outputs changing state), the parallel capacitor can provide the necessary current, which the voltage source may not be able to do because of its parasitic resistance or inductance.
okay, so do you need a resistor in series with your capacitor?
Generally no.
But there are instances where it is used.
One is, when you have a very large capacitor value and you need to limit the inrush current when the voltage source is turned on. In this case you often use a negative-tempco (NTC) resistor to limit resistance during the initial turn-on, but have low loss once it is heated up by current through it.
Another case is that sometimes an RC can be more effective at suppressing noise than a simple capacitor, because the capacitor itself cannot take energy out of the circuit --- it can only store it for later. The series resistor can actually dissipate energy from the circuit.
OUT is simply a label so that the simulator can assign a voltage at that point. It is not part of a charging/discharging circuit.
The basic rule of capacitor charging is that you cannot instantly change the voltage across a capacitor (unlike a resistor). The capacitor in your circuit starts off with no energy and has 0V across it. So OUT will show as 0V. On the rising edge of the input the full voltage of the pulse appears across the 100R resistor. If the step voltage is V then the initial (charging) current will be V/100 amps. As the energy stored in the capacitor increases the voltage across it will increase (Vc). This reduces the size of the current (V - Vc)/100 amps. It is this increase in capacitor voltage that produces the characteristic exponential charging curve.
It will take ONE TIME CONSTANT (C x R) to reach about 67% of the final value.
When the input pulse returns to 0V the capacitor will start to discharge through the 100R resistor.
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Best Answer
I don't know what the capacitor likes or doesn't like, but I think your reasoning is backwards.
First, we usually say "the voltage across an ideal capacitor cannot change instantly" rather than say what the capacitor likes or doesn't like (we might say the capacitor doesn't "like" having a voltage higher than its WV rating across it, or a current greater than its ripple current rating through it, because those things will damage the capacitor).
Second, the reason for the voltage not changing instantly is the capacitor's defining relation:
$$V = \frac{Q}{C}$$
This means that in order to change the voltage (V) instantly, you'd have to change the separated charge (Q) instantly. That would require delivering an infinite current through the capacitor (if only for an instant).
Similarly, the reason the capacitor "acts as a short circuit" for short time periods in a transient analysis is again that its voltage waveform can't be discontinuous without its current waveform including singularities, not the other way around.
Remember
$$V=\frac{Q}{C}$$
Another thing this means is that for the voltage to change quickly (as in a high frequency AC signal), the charge must be moved quickly in and out of the capacitor plates.
Charge moving quickly means large currents.
And large currents through it producing only small voltage changes is exactly what it means to say that a component has a low impedance. When the impedance of the capacitor is much lower than that of other components in our circuit, then we say the capacitor acts like a short circuit.