Why don't voltages add up numerically in an LCR AC circuit?

  • #1
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The attempt at a solution
(Assuming the components are in series of this order: resistor, inductor, capacitor

So I know that the phases are not in sync, that as the voltage oscillates the voltage across each component in the circuit is different.

At the peak voltage aka peak current the voltage across the inductor will be at it's max because it's resisting change in voltage and therefore creates an opposing voltage with an opposing current. When the V starts declining it will then try to keep the current going and produce a voltage the other way

So this means that as it declines and goes 'negative', since the capacitor is attached directly to the power supply from the other end it's voltage will increase as the inductors voltage starts to decrease?

Because I don't know how the voltage of the capacitor really works in relation to the inductor. I know the V of the resistor because it's IR.

I just know that the voltages add up in vectors, which I don't understand the reason of.
 

Answers and Replies

  • #2
BvU
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I just know that the voltages add up in vectors, which I don't understand the reason of.
Because you need two things before you can add two sine waves: amplitude and phase

try to read the treatise 'nothing imaginary about complex numbers' by @LCKurtz on his site (even though it is aimed a teachers, it is an excellent introduction to this funny world of not-so-imaginary numbers)
 
  • #3
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The best way to see this is to connect an oscilloscope across each element. The differences in phase will be apparent.
 
  • #4
Chandra Prayaga
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I assume you are talking of a series AC LCR circuit.
Some points which need to be very carefully understood.
When the components are in series, the order in which you connected them does not matter. The fact that the capacitor is "attached directly to the power supply" has no significance. Wherever the capacitor is connected, the voltage across it will behave the same way.

At the peak voltage aka peak current the voltage across the inductor will be at it's max..
This sentence does not make much sense. Peak voltage is not aka peak current. They are two different things. Also, do you mean that at the instant the voltage of the power supply is maximum, the voltage across the inductor is also maximum? If you do, that is wrong.
 
  • #5
vela
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Since you're dealing with a series circuit, the same current flows through each element. Use that as your common reference point. For simplicity, let's say that ##I(t) = I_0 \sin(\omega t)##. From the relationship between voltage and current for each element, figure out what ##V_\text{L}##, ##V_\text{C}##, and ##V_\text{R}## are. In particular, what is the phase difference between each voltage and the current ##I##?
 
  • #6
CWatters
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The instantaneous voltages do "add up". What I mean is that at any instant in time Kirchoff's voltage law can be applied to the circuit. That's does not mean you can add up the _peak_ voltage across each component and get zero.


Consider a pendulum. As it swings back and forth it trades PE for KE and back again. At any position or instant of time the PE and KE sum to a constant value. However you cannot add the peak KE to the peak PE and claim to have double because those never occur at the same time.
 
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