Is there always a phase shift between current and voltage in an RLC circuit?

AI Thread Summary
In an RLC circuit with an AC supply, the current will always be phase-shifted compared to the voltage if the impedance has an imaginary component or if the circuit is not at resonance. The phase angle α is determined by the impedance, expressed as Z = |Z|*eiα, where α = atan[im(Z)/re(Z)]. This phase shift is crucial for understanding the relationship between voltage and current in AC circuits. The phasor diagram visually represents these relationships, confirming the existence of the phase shift. Therefore, as long as the conditions mentioned are met, the phase shift will always be present.
Nikitin
Messages
734
Reaction score
27
Hey. Say you have an RLC circuit with an AC supply. Then I=ei(ωt-α). My question is: will α always exist, ie will the current always be phase-shifted compared to the voltage, as long as the impedance has an imaginary component or if the circuit is not at resonance? Does Z = |Z|*e, where α = atan[im(Z)/re(Z)] ?
 
Last edited:
Physics news on Phys.org
Does the phasor diagram answer your question?

rlc.gif

Source: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rlcser.html
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top