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Here, in an LCR circuit, when the current is maximum, it will be maximum (and of the same value to maintain the continuity of current) in all the three components, i.e, resistor R, conductor C and inductor I.

Now, why can't we calculate the net voltage by simply adding the voltage drop or raise normally (±). Why to go for a phasor method like this? :-

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And yes, the results are different. Why? Where am I going wrong?

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2 Answers 2

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EDIT#2: Here: http://www.animations.physics.unsw.edu.au/jw/images/AC_files/resonancev(t).gif is an image of the waveforms in this circuit.

You can't simply add the magnitudes of voltage drops across each element to get the maximum voltage. I am not entirely sure, but it looks like in your first equations you are simply writing the magnitudes of VC, VL and VR. You have to incorporate the phase of each. That's why your second method looks correct to me.

Think of it this way. If you are forming a right triangle that has sides, a, b, c such that:

$$ c^2 = a^2 + b^2 $$

If we look at sides a and b as vectors from the origin then we can't simply say that the sum of the magnitudes of a and b is c.

$$ c \ne a + b $$

We can say, however, that:

$$ \vec{c} = \vec{a} + \vec{b} $$

Use this analogy to think about your circuit. Inductors and capacitor have voltage phasors with angles:

$$ \vec{V_L} = IZ = j(I\omega L) = I\omega L\angle 90^\circ $$ $$ \vec{V_C} = IZ = \frac{I}{j\omega C} = -j\frac{I}{\omega C} = \frac{I}{\omega C}\angle{-}90^\circ $$ $$ \vec{V_R} = IZ = IR $$

Then we can get the vector sum, and take its magnitude:

$$ \vec{V_{total}} = \vec{V_L} + \vec{V_C} + \vec{V_R} $$ $$ \vec{\left\lvert V_{total} \right\lvert} = \left\lvert \vec{V_L} + \vec{V_C} + \vec{V_R} \right\lvert $$

In short, the angles are important and because part of the phasor sum.

EDIT: added vector notation for clarity with phasors

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  • $\begingroup$ Okay, almost getting it. But it still bugs me that at the instant of maximum current, the current flowing throught each component will be maximum and hence the voltage (?). Shouldn't we then add them simply? From a perception where I just measure the voltage across each component using voltmeter at that very instant of maximum current and then as a layman, add them simply. $\endgroup$
    – MrObjectOriented
    Commented Oct 21, 2016 at 19:53
  • $\begingroup$ It has to do with the fact that these are sinusoidal voltages (that assumption is implicit in using phasors). Each of the sinusoidal voltage drops produced by each element has different phase shifts. 0, 90 and -90 degrees for the resistor, inductor and capacitor respectively. These phase differences mean they peak at different points. $\endgroup$
    – michael b
    Commented Oct 21, 2016 at 19:55
  • $\begingroup$ I completely get what you are saying of phasors. What I am doing is looking it at another way. Just see like this. You know nothing of phasors. You attach three voltmeters, one each with the resistance, one with inductor and one with the capacitor in parallel. Now at the instant when current flowing through the circuit is maximum, you note the voltage drops across the three elements at that instant and just add them (Kirchhoff's voltage law) to get the net voltage. I hope this is the 'net voltage' we are calculating across the circuit at that instant using phasors? $\endgroup$
    – MrObjectOriented
    Commented Oct 21, 2016 at 20:03
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    $\begingroup$ When the current is at a maximum is not the same instant that the voltage across everything is at a maximum. In fact, only the voltage across the resistor will be at a peak at the same time as the current. When you convert the phasors back to the time domain, you will get a voltage that is a function of time. This equation will represent the circuit's instantaneous voltage. $\endgroup$
    – michael b
    Commented Oct 21, 2016 at 20:11
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    $\begingroup$ They are shifted by 90 degrees either way, so they won't be at pick. Look here: animations.physics.unsw.edu.au/jw/images/AC_files/… , it's the the voltage waveforms and keep in mind that when VR is at peak, current is at peak. $\endgroup$
    – michael b
    Commented Oct 21, 2016 at 20:30
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In this series connected RLC resonant circuit the maximum current occurs at the resonance condition $$\omega L=1/\omega C$$ In this case, the impedance of the inductor-capacitor series connection becomes zero because the voltage drops over the capacitor and the inductor have opposite phase summing up to zero voltage. At resonance, the current through the RLC connection is given by $$I=V_0/R$$ and the voltage drop over the capacitance is $$V_C=-j\frac{I}{\omega C}=-j\frac{V_0}{\omega RC}$$ and over the inductance $$V_L=jI\omega L=j\omega L\frac{V_0}{R}$$ It can be seen that for $$\omega RC=R/\omega L<1$$ the amplitudes of the voltage drops over the capacitance and the inductance, $V_C$ and $V_L$, can easily become much larger than the applied voltage $V_0$. This is a common phenomenon in resonant circuits which can lead to the destruction (breakdown) of the circuit elements.

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