Question

Consider an RLC circuit that is driven by an AC source with an adjustable frequency. Is possible for the maximum voltage drop across the capacitor to exceed the maximum voltage? Explain in 100 words or less.

I know the answer is yes, but can you explain this conceptually, not mathmatically? thanks.

Answer 1

Consider an RLC series circuit

If resonance occurs at a frequency f and where the total voltage (the voltage across the source) is V and the current is I

a. The voltage drop across the resistor, labelled as V(r) will be in phase with I and of magnitude IR

b. The voltage drop across the inductor, labelled as V(l), will be leading I by π/2 and of magnitude V(l)=ILω, where ω = 2πf

c. The voltage drop across the capacitor, labelled as V(c), will be lagging behind I by π/2 and of magnitude V(c)=I/Cω

d. At resonance V(l) = - V(c) so in magnitude V(l) = V(c) and hence combining the equations from b. and c. above we get ILω = I/Cω, so cancelling I from both sides you get Lω = 1/Cω which implies that ω squared = 1/LC so ω = √(1/LC)

e. The net voltage V = V(r) + V(l) + V(c) , at resonance V(l) and V(c) cancel each other, because V(l) = - V(c), so V = V(r) (The source voltage becomes equal to the voltage across R)

f. The magnitude of the net voltage at resonance is hence V= V(r) = IR

Now back to the question which is about comparing V(l) to V at resonance:

V(l)= ILω = IL√(1/LC) (from b. above ), So V(l) = I√(L/C)

but we know that V = IR (from f. above)

So V(l)/V = I√(L/C)/IR = √(L/C)/R

Hence if √(L/C) > R, the voltage V(l) across the inductor, at resonance, will be larger than the source voltage V.

This of course does not mean that V(l) will always greater than V at resonance, but it does mean that V(l) compares to V depending on the relative values of √(L/C) and R.