# Series RLC and Parallel RLC circuits

Summary: Series RLC and Parallel RLC circuits

How can the voltage across a capacitor or inductor in a series RLC circuit be greater than the applied AC source voltage? The formula suggest that either can be larger than the source voltage but I still find it counter intuitive. Also for Parallel RLC, how can the current of anyone branch can be larger than the AC source current?

Please explain to me the mechanism as to how these phenomenon are possible. THANK YOU!

BvU
Homework Helper
With a continuing series of small pushes you can let a swing go very high
In the circuits there can be a resonance when the charge piles up in the C, then flows back through the L nicely at the right time -- the L tries to 'maintain the current', thus charging C on the other side, etc.

Resonance is the key word here.

• Klystron and berkeman
With a continuing series of small pushes you can let a swing go very high
In the circuits there can be a resonance when the charge piles up in the C, then flows back through the L nicely at the right time -- the L tries to 'maintain the current', thus charging C on the other side, etc.

Resonance is the key word here.
What do you mean by "series of small pushes " and "flows back through the L nicely at the right time"?
Please bear with me English is my second language.

Borek
Mentor
Not exactly a typical RC circuit, but think in terms of an induction coil - you feed it with a low (pulsing) voltage, you get high voltage. That's basically the same principle, energy stored in the magnetic field induces the voltage which can be much higher than the voltage used to produce the filed initially.

BvU
Homework Helper
What do you mean by "series of small pushes " and "flows back through the L nicely at the right time"?
Please bear with me English is my second language. https://en.wikipedia.org/wiki/LC_circuit

or else look here under 'Basic LC Oscillator Tank Circuit'

Wonderful graphics here

Tom.G
Here is a simple text version of what happens.

Consider a parallel RLC circuit driven at it's resonant frequency. On the first half cycle the capacitor charges up to some voltage. On the next half cycle the driving voltage is the opposite polarity of the capacitor. That makes two voltage sources in series. With the higher voltage and the same circuit resistance the current thru the capacitor will be higher.

What it comes down to is that there is a circulating current with the capacitor and inductor exchanging their stored energy with each other. This is in addition to the energy supplied by the source during any particular cycle. That's why you measure more current in the components than is coming from the supply.

Hope this helps.

Cheers,
Tom

Here is a simple text version of what happens.

Consider a parallel RLC circuit driven at it's resonant frequency. On the first half cycle the capacitor charges up to some voltage. On the next half cycle the driving voltage is the opposite polarity of the capacitor. That makes two voltage sources in series. With the higher voltage and the same circuit resistance the current thru the capacitor will be higher.

What it comes down to is that there is a circulating current with the capacitor and inductor exchanging their stored energy with each other. This is in addition to the energy supplied by the source during any particular cycle. That's why you measure more current in the components than is coming from the supply.

Hope this helps.

Cheers,
Tom
Hello! I'm still confused. I don't quite get this part "On the next half cycle the driving voltage is the opposite polarity of the capacitor. That makes two voltage sources in series". I thought we're taking about parallel RLC, why the series voltage soruces occurs? and what about the inductor? What is it doing during this time?

anorlunda
Staff Emeritus
Hello! I'm still confused.
Did you watch this video that @BvU mentioned?

Did you watch this video that @BvU mentioned?
I watched it already but still find it difficult to understand as to why these phenomenon occurs.
For example in parallel RLC when driven at resonance the total impedance is just R which is the maximum then current drawn from supply will be minimum and by ohm's law it is just Vs/R. The reactances of Capacitor and inductor basically canceled each other out and can be replaced by an open circuit. So how is it possible that they can have current flowing through them greater than the source current if they are likened to an open circuit?

anorlunda
Staff Emeritus
The reactances of Capacitor and inductor basically canceled each other out and can be replaced by an open circuit.
That is true only for averages over an integral number of whole cycles. If you look within a cycle, they act at different times. Instead of AC circuit analysis, you should be using DC analysis, with LdI/dt and CdV/dt which gives you differential equations, not algebraic equations.

kimbyd
Gold Member
I watched it already but still find it difficult to understand as to why these phenomenon occurs.
For example in parallel RLC when driven at resonance the total impedance is just R which is the maximum then current drawn from supply will be minimum and by ohm's law it is just Vs/R. The reactances of Capacitor and inductor basically canceled each other out and can be replaced by an open circuit. So how is it possible that they can have current flowing through them greater than the source current if they are likened to an open circuit?
Because the flowing current is in a cycle, while the source gives the cycle a periodic kick.

If you look at how the voltage changes over time and start with the system with the power source disconnected, it will start with no voltage differences and no current. On the first cycle after connecting the power source, it causes a bit of power to flow through. Some of that power goes into a cycle through the capacitor and inductor: the capacitor resists changes in voltage, maintaining a power differential that pushes current through the inductor, while the inductor resists changes in current, pushing current onto the capacitor plates and maintaining a voltage differential.

So the LC part of the circuit tends to oscillate, with the capacitor driving the inductor and vice versa. The resistor siphons off a bit of the voltage in each cycle, because both the capacitor and inductor are also pushing back against one another's behavior.

What happens if you leave the oscillating source connected? If its frequency is right for the circuit, then every cycle it will enhance the cycle, creating larger and larger swings in voltage and current. Eventually the resistance of the LRC circuit will prevent further increases in voltage, but it's not hard to build a circuit that will have 100x the voltage oscillations of its input.

Note that you aren't creating new energy here at all: you're just putting a little bit of energy into the oscillating system at a time. If there's no release, it will eventually build up.

Because the flowing current is in a cycle, while the source gives the cycle a periodic kick.

If you look at how the voltage changes over time and start with the system with the power source disconnected, it will start with no voltage differences and no current. On the first cycle after connecting the power source, it causes a bit of power to flow through. Some of that power goes into a cycle through the capacitor and inductor: the capacitor resists changes in voltage, maintaining a power differential that pushes current through the inductor, while the inductor resists changes in current, pushing current onto the capacitor plates and maintaining a voltage differential.

So the LC part of the circuit tends to oscillate, with the capacitor driving the inductor and vice versa. The resistor siphons off a bit of the voltage in each cycle, because both the capacitor and inductor are also pushing back against one another's behavior.

What happens if you leave the oscillating source connected? If its frequency is right for the circuit, then every cycle it will enhance the cycle, creating larger and larger swings in voltage and current. Eventually the resistance of the LRC circuit will prevent further increases in voltage, but it's not hard to build a circuit that will have 100x the voltage oscillations of its input.

Note that you aren't creating new energy here at all: you're just putting a little bit of energy into the oscillating system at a time. If there's no release, it will eventually build up.
Hello,

"then every cycle it will enhance the cycle" what cylce are you referring to in this statement?
" creating larger and larger swings in voltage and current" in RLC parallel configuration you apply the voltage source in parallel to the components how can the swing in voltage be larger? Is not it the same all through out?
Please bear with me. I really want to understand this concept.

Thank You.

kimbyd
Gold Member
Hello,

"then every cycle it will enhance the cycle" what cylce are you referring to in this statement?
" creating larger and larger swings in voltage and current" in RLC parallel configuration you apply the voltage source in parallel to the components how can the swing in voltage be larger? Is not it the same all through out?
Please bear with me. I really want to understand this concept.

Thank You.
When you create a current pulse through an LRC circuit (think connecting a battery for an instant then disconnecting it), it starts a cycle of current flow through the inductor and capacitor, with the resistor siphoning off some of the excess to make everything balance. Voltage will start to cycle up and down, and current flow will cycle back and forth in a loop through the circuit. Over time, the resistor siphons off the energy and the cycle dies.

But if you have a varying voltage source that has the same periodicity as the circuit's fundamental frequency, then every "kick" of the voltage source enhances the cycle (until the energy siphoned off from the resistor equals the energy input into the system each voltage cycle).

vela
Staff Emeritus
Homework Helper
The reactances of Capacitor and inductor basically canceled each other out and can be replaced by an open circuit. So how is it possible that they can have current flowing through them greater than the source current if they are likened to an open circuit?
Your mistake is in thinking the inductor and capacitor can be replaced by an open circuit. This is wrong because there is still current flowing through the two elements.

Assume that the driving voltage is given by ##v(t) = V_0 \sin\omega t## and calculate the current through each element. You should find that when the supply voltage oscillates at the frequency ##1/\sqrt{LC}##, the current supplied by the capacitor is equal to the current flowing into the inductor, so no additional current needs to be supplied by the power supply for these two elements. The voltage source only needs to supply current for the resistor, so it sees an impedance ##R##.

At resonance, the amplitude of the current supplied by the source is ##V_0/R##. What determines the amplitude of the current oscillation through the inductor and capacitor? Is there any reason to believe that one amplitude must always be smaller than the other?

Another question you might consider: Suppose you have an LC circuit (with no external power supply). At ##t=0##, no current is flowing, and the capacitor has a charge ##q_0##. The circuit is going to oscillate, right? The supplied current is obviously 0, yet the current in the circuit is not 0. What's going on here?

kimbyd
Gold Member
To see how the cycle operates, consider what happens the second the battery is connected. Let's imagine it's +3V. When it's connected, all three components have +3V across them. The 3V/R current flows through the resistor. No current flows through the inductor (because inductors resist changes to current). Lots of current instantly flows through the capacitor, while charge starts to build up within the capacitor creating a voltage differential between its plates.

Then the battery is disconnected. But there's a capacitor sitting there that has built up some voltage from the time it was connected. In general it will be something less than +3V, as the battery was only connected for a moment. Let's imagine it builds up +1V. So, +1V/R current flows through the resistor. As the inductor doesn't block current flow (it only resists changes to it), some current is flowing through the inductor by now. Meanwhile, the capacitor is discharging the voltage stored in it.

After a little while, the capacitor becomes entirely drained. Current stops flowing through the resistor. But the inductor is still pushing current through. This forces current through the capacitor, which in turn starts building up charge in the opposite direction. Perhaps -0.5V or so. Current flows back the other way.

So the current flow sloshes back and forth until all the energy is siphoned off. The timing of the cycle depends upon the resistance, inductance, and capacitance of the three components.

.Scott
Homework Helper
This can happen with any oscillating system.
When you were young, you may have learned to "pump" yourself on a playground swing.
To do this, you don't need to apply all the power in one swing - just a little at the right time on each swing. And it doesn't work if you apply a small steady push. It has to be a periodic push - timed to the swinging. When you do it right, each swing is little higher than the one before.

• SammyS and DaveE