Understanding L-C-R Circuits: Can Voltages Exceed the Source? | Explained

[Reshma]

I have an L-C-R circuit with an ac source. $V_L$, $V_R$ & $V_c$ are the peak voltages across the inductor, resistor and capacitor. V is the source voltage. Can the voltage across the capacitor and inductor exceed the source voltage? Can this be true for the resistor?

My answer:

Well, the formula relating the voltages is:
$$V = \sqrt{V_R^2 + (V_L - V_C)^2}$$

So the voltages across the inductor and capacitor can exceed the source voltage but the voltage across the resistor must always be less than or equal to the source.

Now is my answer correct? Is there any furthur explanation required here?

 

[marlon]

Well, just look at the equation mathematically :

$$V^2 = V_{R}^2 + (V_{L}-V_{C})^2$$

So you have that :

$$V^2 >= V_{R}^2$$
$$V^2 >= (V_{L}-V_{C})^2$$

Now, when taking the square root, just divide the problem in two cases:
1) V is positive so that $$\sqrt{V^2} = V$$

2)1) V is negative so that $$\sqrt{V^2} = -V'$$, and V' is positive.

For example you can have :
$$(-3)^2 > (-2)^2$$ ---> (-3) < (-2)
$$(3)^2 > (2)^2$$ ---> (3) > (2)

You see ?

marlon

 

[Reshma]

Well, just look at the equation mathematically :

$$V^2 = V_{R}^2 + (V_{L}-V_{C})^2$$

So you have that :

$$V^2 >= V_{R}^2$$
$$V^2 >= (V_{L}-V_{C})^2$$

Now, when taking the square root, just divide the problem in two cases:
1) V is positive so that $$\sqrt{V^2} = V$$

2)1) V is negative so that $$\sqrt{V^2} = -V'$$, and V' is positive.

For example you can have :
$$(-3)^2 > (-2)^2$$ ---> (-3) < (-2)
$$(3)^2 > (2)^2$$ ---> (3) > (2)

You see ?

marlon

Thanks for the reply.

So, if V is positive:
$V>=V_R$
$V>=V_L - V_C$

if V is negative:
$V<=V_R$
$V<=V_L - V_C$

But I don't infer much from this. One thing is for sure, either V_L or V_C can exceed V, since we consider their difference. But what about V_R?

There is also another inequality, I wonder if that can be applied here.
$$V \neq V_R + V_L + V_C$$

 

[Curious3141]

Thanks for the reply.

So, if V is positive:
$V>=V_R$
$V>=V_L - V_C$

if V is negative:
$V<=V_R$
$V<=V_L - V_C$

But I don't infer much from this. One thing is for sure, either V_L or V_C can exceed V, since we consider their difference. But what about V_R?

There is also another inequality, I wonder if that can be applied here.
$$V \neq V_R + V_L + V_C$$

Reshma, I think there's a little confusion here. Let's consider the series R, L, C circuit a little more carefully.

The equation $$V^2 = (V_L - V_C)^2 + V_R^2$$ pertains to peak voltages. Each of the voltages should be assumed to take only positive values, since they are all peak values (amplitudes of phase-shifted sinusoidal functions).

To avoid confusion, let's rewrite that as $$V_{s,peak}^2 = (V_{L,peak} - V_{C,peak})^2 + V_{R,peak}^2$$ where I introduced the suffix 's' to denote supply.

In the above, certainly both $$V_{L,peak}$$ and $$V_{C,peak}$$ can exceed $$V_{s,peak}$$. $$V_{R,peak}$$ can never exceed $$V_{s,peak}$$.

I hope all that's quite clear to you by now.

Now let's consider instantaneous voltages across each component, representing them as functions of time, viz. $$V_{L}(t), V_{C}(t), V_{R}(t)$$. The supply will be represented as $$V_{s}(t)$$.

When we're considering instantaneous voltages, the following equality always holds true :

$$V_{s}(t) = V_{L}(t) + V_{C}(t) + V_{R}(t)$$

and you can prove that with a little complex number or trigonometric manipulation. That's simply Kirchoff's first law, there's no reason it fails in reactive a.c. circuits. But what you must keep in mind is that each of those instantaneous voltages is a signed quantity. The voltage over the inductor may be of opposite sign to that across the capacitor, and you should take care when adding them up algebraically. In addition, note that any of the instantaneous voltages (including that across the resistor) may be greater (in magnitude) than the corresponding instantaneous supply voltage; however, the instantaneous voltage across the resistor can never be greater than the peak supply voltage. In contrast, the instantaneous voltages across either the inductor or the capacitor may actually be greater in magnitude than the peak supply voltage.

To put it simply, when considering peak voltages, the inequality you mentioned can hold; however, when considering signed instantaneous voltages, you will get an equality correspoding to Kirchoff's first law.

 

[Reshma]

Wow, thank you very much, Curious3141. I guess I was confusing the peak voltages with the instantaneous voltages .
So from the loop rule, at any instant the sum of the voltages across the resistor, capacitor and inductor equals the applied emf.
But, when plotting the phasors of each of the voltages, we consider only the peak voltages. $V_{s,peak}$ must be equal to the vector sum of the three voltage phasors $V_{R,peak} V_{L,peak}V_{C,peak}$. Since phasors $V_{C,peak}$ & $V_{L,peak}$ have opposite directions they can be combined to form a single phasor $V_{L,peak}-V_{C,peak}$.

 

[Curious3141]

Wow, thank you very much, Curious3141. I guess I was confusing the peak voltages with the instantaneous voltages .
So from the loop rule, at any instant the sum of the voltages across the resistor, capacitor and inductor equals the applied emf.
But, when plotting the phasors of each of the voltages, we consider only the peak voltages. $V_{s,peak}$ must be equal to the vector sum of the three voltage phasors $V_{R,peak} V_{L,peak}V_{C,peak}$. Since phasors $V_{C,peak}$ & $V_{L,peak}$ have opposite directions they can be combined to form a single phasor $V_{L,peak}-V_{C,peak}$.

Yes, the usual convention when plotting a phase diagram is to let the length of the phasors represent the peak values.