Parallel resistor-capacitor circuits

Using the same value components in our series example circuit, we will connect them in parallel and see what happens: (Figure below)

Parallel R-C circuit.
Because the power source has the same frequency as the series example circuit, and the resistor and capacitor both have the same values of resistance and capacitance, respectively, they must also have the same values of impedance. So, we can begin our analysis table with the same “given” values:

This being a parallel circuit now, we know that voltage is shared equally by all components, so we can place the figure for total voltage (10 volts ∠ 0o) in all the columns:

Now we can apply Ohm's Law (I=E/Z) vertically to two columns in the table, calculating current through the resistor and current through the capacitor:

Just as with DC circuits, branch currents in a parallel AC circuit add up to form the total current (Kirchhoff's Current Law again):

Finally, total impedance can be calculated by using Ohm's Law (Z=E/I) vertically in the “Total” column. As we saw in the AC inductance chapter, parallel impedance can also be calculated by using a reciprocal formula identical to that used in calculating parallel resistances. It is noteworthy to mention that this parallel impedance rule holds true regardless of the kind of impedances placed in parallel. In other words, it doesn't matter if we're calculating a circuit composed of parallel resistors, parallel inductors, parallel capacitors, or some combination thereof: in the form of impedances (Z), all the terms are common and can be applied uniformly to the same formula. Once again, the parallel impedance formula looks like this:

The only drawback to using this equation is the significant amount of work required to work it out, especially without the assistance of a calculator capable of manipulating complex quantities. Regardless of how we calculate total impedance for our parallel circuit (either Ohm's Law or the reciprocal formula), we will arrive at the same figure:

* REVIEW: * Impedances (Z) are managed just like resistances (R) in parallel circuit analysis: parallel impedances diminish to form the total impedance, using the reciprocal formula. Just be sure to perform all calculations in complex (not scalar) form! ZTotal = 1/(1/Z1 + 1/Z2 + . . . 1/Zn) * Ohm's Law for AC circuits: E = IZ ; I = E/Z ; Z = E/I * When resistors and capacitors are mixed together in parallel circuits (just as in series circuits), the total impedance will have a phase angle somewhere between 0o and -90o. The circuit current will have a phase angle somewhere between 0o and +90o. * Parallel AC circuits exhibit the same fundamental properties as parallel DC circuits: voltage is uniform throughout the circuit, branch currents add to form the total current, and impedances diminish (through the reciprocal formula) to form the total impedance.

Well I have read that capacitor acts as a short circuit when it is first connected to a source and large surge or inrush current can flow through the circuit.This current may damage the capacitor or other components So to limit this current, a resistor is connected in series with a capacitor.I studied the operation of series RC circuit.Now when i googled it, I found RC parallel circuit.Something like this belowSo in the above circuit I expect same voltage appears across capacitor and resistor.Since there is no current limiting resistor in series with capacitor. I expect the surge current exists in the parallel branch of capacitor and these may damage the capacitor but the author no where mentioned to put a resistor in series with a capacitor.I expect the practical RC parallel circuit is like I have shown you below. To limit the surge current its a must to put a series resistor with a capacitor.So kindly let me know is it necessary to put a series resistor with a capacitor.Let me know which one of the above circuit is right.Could someone explain me If the source is DC in a RC parallel circuit does we need a series resistor with a capacitor or not and if the source is AC in RC parallel circuit does we need a series resistor with capacitor or not.capacitor shareimprove this question | edited Feb 8 '13 at 8:11Federico Russo
3,86132984 | asked Feb 8 '13 at 4:21niko
2682311 | | | add a comment |
1 Answer activeoldestvotes up vote10down voteaccepted | 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. |

The Circuit

In the schematic diagram shown to the right, we show a parallel circuit containing an ideal inductance and an ideal capacitance connected in parallel with each other and with an ideal signal voltage source. For consistency, we will use the same example values that we used when examining the series LC circuit. Thus, * VAC = 10 vrms. * f = 1 MHz. (ω = 6283185.3 rad/sec) * L = 150 µh. (XL = 942.4778 Ω) * C = 220 pf. (XC = 723.43156 Ω)
According to Ohm's Law: iL = vL/XL = 10/942.4778 = 0.01061033 = 10.61033 mA. iC = vC/XC = 10/723.43156 = 0.013823008 = 13.823008 mA.
If we measure the current provided by the source, we find that it is 3.2126777 mA — the difference between iL and iC.
The question to ask about this circuit, then, is, "Where does the extra current in both L and C come from, and where does it go?"

The Vectors

The vectors that apply to this circuit imply the answer, as shown to the right. Here, the voltage is the same everywhere in a parallel circuit, so we use it as the reference. There is no resistance, so we have no current component in phase with the applied voltage.
We already know that current lags voltage by 90° in an inductance, so we draw the vector for iL at -90°. Similarly, we know that current leads voltage by 90° in a capacitance. Therefore, we draw the vector for iC at +90°.
Combining these two opposed vectors, we note that the vector sum is in fact the difference between the two vectors. This matches the measured current drawn from the source.
The remaining current in L and C represents energy that was obtained from the source when it was first turned on. This energy, and the current it produces, simply gets transferred back and forth between the inductor and the capacitor.
If we begin at a voltage peak, C is fully charged. Since current is 90° out of phase with voltage, the current at this instant is zero. But C now discharges through L, causing voltage to decrease as current increases. When C is fully discharged, voltage is zero and current through L is at its peak. This current has caused the magnetic field surrounding L to increase to a maximum value. This completes ¼ cycle.
The second quarter-cycle sees the magnetic field collapsing as it tries to maintain the current flowing through L. This current now charges C, but with the opposite polarity from the original charge. As current drops to zero and the voltage on C reaches its peak, the second ¼ cycle is complete.
The other half of the cycle sees the same behavior, except that the current flows through L in the opposite direction, so the magnetic field likewise is in the opposite direction from before. At the conclusion of the second half-cycle, C is once again charged to the same voltage at which it started, with the same polarity. Now, a new cycle begins and repeats the actions of the old one.

The Mathematics
The calculation for the combined impedance of L and C is the standard product-over-sum calculation for any two impedances in parallel, keeping in mind that we must include our "j" factor to account for the phase shifts in both components. Thus, Z | = | (jXL)(-jXC) | | | | | | (jXL) + (-jXC) | | | = | -j²(XL)(XC) | | | | | | j(XL - XC) | | | = | -j(XL)(XC) | | | | | | (XL - XC) | |
This equation tells us two things about the parallel combination of L and C: 1. The overall phase shift between voltage and current will be governed by the component with the lower reactance. This is reasonable because that will be the component carrying the greater amount of current. 2. The impedance of the parallel combination can be higher than either reactance alone. This is because of the opposed phase shifts in current through L and C, forcing the denominator of the fraction to be the difference between the two reactances, rather than the sum of them.
Because the denominator specifies the difference between XL and XC, we have an obvious question: What happens if XL = XC — the condition that will exist at the resonant frequency of this circuit? Clearly there's a problem with a zero in the denominator of a fraction, so we need to find out what actually happens in this case.

When XL = XC
At the resonant frequency of the parallel LC circuit, we know that XL = XC. At this frequency, according to the equation above, the effective impedance of the LC combination should be infinitely large. In fact, this is indeed the case for this theoretical circuit using theoretically ideal components.
The currents flowing through L and C may be determined by Ohm's Law, as we stated earlier on this page. The current drawn from the source is the difference between iL and iC. However, when XL = XC and the same voltage is applied to both components, their currents are equal as well. Therefore the difference is zero, and no current is drawn from the source. This corresponds to an infinite impedance, or an open circuit.
This doesn't mean that no current flows through L and C. Rather, all of the current flowing through these components is simply circulating back and forth between them without involving the source at all. The currents calculated with Ohm's Law still flow through L and C, but remain confined to these two components alone. As a result of this behavior, the parallel LC circuit is often called a "tank" circuit, because it holds this circulating current without releasing it.
There is one other factor to consider when working with an LC tank circuit: the magnitude of the circulating current. We can use many different values of L and C to set any given resonant frequency. Keep in mind that at resonance: XL | = | XC | | ωL | = | 1 | | | | | | ωC | | ω² | = | 1 | | | | | | LC | | ω | = | 1 | | | | | | |
As long as the product L × C remains the same, the resonant frequency is the same. However, if we use a large value of L and a small value of C, their reactances will be high and the amount of current circulating in the tank will be small. If we reverse that and use a low value of L and a high value of C, their reactances will be low and the amount of current circulating in the tank will be much greater. Many applications of this type of circuit depend on the amount of circulating current as well as the resonant frequency, so you need to be aware of this factor. In fact, in real-world circuits that cannot avoid having some resistance (especially in L), it is possible to have such a high circulating current that the energy lost in R (p = i²R) is sufficient to cause L to burn up!