What actually goes on in RLC circuit?

Question

I already have difficulty understanding how each of the RLC circuit elements work individually as it is, and I have no idea how they work simultaneously in an RLC circuit. My schoolbook represents the components' resistance, reactance and inductance as vectors and by some math derive their sum. It then goes on to consider all elements as one, producing a net impedance equal to this derived total. This representation is easy on the mind and facilitates solving the problems but I want to intuitively understand how all these out of phase Ashe components work together and in what progression do they each impact the emf of the circuit. I need someone to kind of describe to me the first few moments of closing an RLC AC source circuit..

Answer 1

If you have a series LCR circuit then at every instant of time the current in each part of the circuit must be the same which is KCL - the conservation of charge.
Let that current be $I=I_o \sin (\omega t)$

Now one must add up all the voltages in the circuit and make the sum zero.
The complication is that although all the voltages will be of the same frequency $\omega$ they will not be in phase with each other.
The voltage across the resistor is easy because it will be in phase with the current ($V_R = RI$) and have a value of $V_R = R I_o \sin(\omega t)$

For the inductor $V_L = L \frac {dI}{dt} \Rightarrow V_l = \omega L I_o \cos (\omega t) = \omega L I_o \sin (\omega t + \frac \pi 2) $
The voltage across the inductor is equal to the rate of change of current passing through it.
When the rate of change of current is a maximum, that is when the current is zero, the voltage is a maximum. There is therefore a $\frac \pi 2$ difference in phase between the current in the circuit and the voltage across the inductor with the current leading the voltage. Whatever the current does, ie $I(t) = 0$ the voltage does a quarter of a period later $V(t+\frac T 4)=0$

For the capacitor $V_C=Q/C.$ The current flowing onto the positive capacitor plate is by definition the rate at which charge is being stored. So the charge Q on the capacitor equals the integral of the current with respect to time i.e.: $$V=\frac{Q}{C}=\frac{1}{C}\int_t{I(t)dt}$$ Therefore $$V_C = -\frac {I_o}{\omega C} \cos (\omega t) = \frac {I_o}{\omega C} \sin (\omega t - \frac \pi 2) $$ In this case the voltage across the capacitor lags the current by $\frac \pi 2$.
For example when the voltage across the capacitor is zero you have a maximum (charging) current flowing and a quarter of a period later the current is zero.

So in the end you have to add three voltages $R I_o \sin(\omega t) + \omega L I_o \sin (\omega t + \frac \pi 2) + \frac {I_o}{\omega C} \sin (\omega t - \frac \pi 2)$ and make their sum equal to the voltage of the supply.

The addition can be done in numerous ways but one relatively straightforward way is to use the idea of phasors.

This web page produced by the University of New South Wales has some nice animations to illustrate the phase differences and the use of phasors.

Answer 2

When you apply an AC source of given frequency to the components R, L and C in series the current through all the components must be the same. The use of vectors are introduced because for the different components the relationship between the voltage across the component and the current through them are different: For R the voltage is in phase with the current, while for L and C the voltages are not in phase with the current. This is because, inductors don't like current changes (when you change the current in an inductor the magnetic field linking it changes and changing magnetic fields generate voltages that oppose the change in current) and capacitors don't like voltage changes across them. To take these things into account you introduce phase differences in the voltages across the components. Phase differences are more easily included mathematically by using complex numbers and complex numbers can (roughly) be treated as having "vector-like" properties.