Question

(a) In an RLC circuit, can the amplitude of the voltage across an inductor be greater than the amplitude of the generator emf? (b) Consider an RLC circuit with E = 10 V, R = 10 Ω, L = 1.0 H, and C = 1.0 μF. Find the amplitude of the voltage across the inductor at resonance.

Answer 1

Assuming the RLC circuit is series resonant, the resonant frequency is found from:
\[\large f=\frac{1}{2\pi \sqrt{LC}}\]
At series resonance, the current is:
\[\large i=\frac{E}{R}=1\ A\]
At series resonance, the voltage across the inductor is:
\[\large v _{l}=i \times X _{l}=1\times2\pi \times f \times L\]

Answer 2

do you know phasor diagrams?

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/phase.html

you can intuit the answer pretty much straight away if you do

@Radar is great at this stuff. @Michele_Laino too

and the French guy who i haven't fanned, but will, and then tag !

the DE is in terms of I(t) or i would just recommend putting that into Wolfram that with the numbers you have given.

Answer 3

that's right! @kropot72

Answer 4

@Vincent-Lyon.Fr

michele: i was thinking more about the idea that you could have a back voltage on the inductor greater that the voltage on the source. i think the easiest way to see round that is using a phasor diagram.

Answer 5

yes! Of course, nevertheless please keep in mind at condition of resonance, the inductive impedance is equal to the capacitive impedance, so we have this situation:

Answer 6

namely the applied voltage has the same phase of the current

Answer 7

cool!

so now we know