# What's the purpose of capacitors in parallel

In my school textbook it is written that the capacitor acts as a filter, that is, it decreases the fluctuations in the potential difference across the load.

But since all the components are connected to each other in parallel the potential difference across them should be same. So there should be no change in potential difference across the load even if a capacitor is connected in parallel.

Can anyone explain this to me?

A capacitor can contain a certain amount of charge for a given voltage:
$$Q = CV$$ When you have more than one capacitor in parallel, they have the same voltage (because they are in parallel), and each stores a certain charge. The total charge (at a given voltage) will be the sum of the charges on all the capacitors.

Now if you have a certain load (for example a resistor in parallel with the capacitors), that load will draw a particular current (charge per unit time). If more charge is stored (because the capacitance is greater), then the voltage will drop less per unit time. This means that if you have a bridge rectifier, like in your diagram, and you have a certain load (not shown in your diagram), then the "ripple" on the power supply will be less if the capacitance is greater.

The basic effect is shown in this diagram:

You can see the AC signal, the (bridge) rectified signal, and the signal after the capacitor (with a certain current being drawn). As the capacitor gets larger, the amount of voltage droop will be smaller (the slope of the green curve will be less if the capacitance is greater as the capacitor can provide more charge / current without the voltage decreasing).

Incidentally, sometimes people will put capacitors of different types in parallel. For example, a large electrolytic capacitor (1000 µF) and a small ceramic capacitor (100 nF). This is done because "real" capacitors have a series inductance - and in the parallel case, the small capacitor (which has a smaller inductance) will be able to respond quickly to rapid changes in current, while the larger capacitance will take care of "longer term" current demands. This is sometimes called "supply decoupling". It's probably outside of the scope of your current question, but a very important principle in electrical engineering.

• If I'm not mistaken, the circuit in the OP after connection of the load would be equivalent to this one. Note that the load is in parallel with the capacitor. How does the capacitor provide anything better than the supply does? Or is it that the supply mustn't be considered ideal voltage source? Jan 18, 2016 at 6:22
• @Ruslan - the supply is AC (you know this because of the transformer). Jan 18, 2016 at 6:43
• Yes, and the output is a fixed function $V(t)$. As the time passes, the supply forces exact values of voltage on its output (unless it's not ideal and has a series resistance), and due to direct parallel connection, same values of voltage on input of capacitor and load. Jan 18, 2016 at 6:59
• @Ruslan - I don't understand what you are saying. The green line in my plot shows that the voltage is NOT a fixed value - it changes over time. During the peak of the cycle, the input voltage of the bridge is high enough to charge the capacitor; but for most of the cycle, the diodes are reverse biased and the capacitor provides the charge / current / voltage for the output. Thus the capacitor is ESSENTIAL to provide a reasonably constant output voltage of the loaded bridge rectifier. Jan 18, 2016 at 7:19
• I don't mean fixed value as constant of time — I mean a fixed function from time to voltage. Jan 18, 2016 at 9:16

Your misunderstanding comes from assuming that the load side is always shorted to the transformer output. This is not the case. The voltage drop across the capacitor causes all of the diodes to turn off when the voltage falls below the peak so that the load is no longer shorted to the transformer and thus need not have the same voltage as it. Let me explain with some diagrams. Assume we have ideal diodes. Consider an input sine wave (on the transformer secondary terminals).

At $t=0$, the secondary voltage is zero, the capacitor voltage is zero and the load voltage is zero. All the diodes are shorted. No current is flowing. Everything is zero basically.

Now consider a time $t=0+dt$. The voltage drop across two of the diodes is now just barely positive, causing the diodes to short. The other two diodes are the opposite: they are open circuits now.

Just as you would expect, this would cause the capacitor and the load to both be at the voltage of the input. This continues until the input voltage reaches its positive peak at time $t=T$.

Now at time $t=T+dt$, the input voltage falls infinitesimally below its peak, and since the maximum input voltage is stored in the capacitor, the diodes which were previously shorted now have a slightly higher voltage at their negative terminal, causing them to become open circuits. This means the load is no longer shorted to the source and need not have the same voltage as it. The other diodes remain open circuits also for the same reason. So what results is this:

This means that the capacitor will now just discharge through the load. Now the input voltage is on its negative cycle. Once the absolute value of the input voltage just barely exceeds the capacitor's voltage, the other two diodes will short - recharging the capacitor until the input reaches its peak:

So in a summary, your analysis is correct - for only part of the input cycle:

For section A of the input cycle, two of the diodes are shorted like in the first image above. As you'd expect, the load voltage is thus the same as the rectified input voltage. However - and this is where you go wrong - in part B, all the diodes are open circuited, so only the capacitor determines the load voltage. In part C, the other two diodes are shorted, and the load voltage is again the same as the input rectified voltage.

Things are a little bit different in the non-ideal case, with forward diode drops and so on, but overall it's the same principle.

(Note: questions like these are better submitted to the Electronics Stack Exchange).