# Problem in understanding analytical solution of LCR circuit

1. Sep 7, 2014

### ovais

Hi all,

I in my text they first did a phasor-diagram solution to a series LCR circuit and brought Z= under root of (R^2 +(Xc^2-XL^2)).

After this they use a differential equation for series LCR circuit and actually did not solve such hard two degree differential equation, rather they assume the solution to be q= q• sin(wt+€) and taking its first and second derivatives and used them in orignal differential equation. Then they divide it by Z= square root of(R^2+ (Xc^2-XL^2)) and tan¥=R/Z so that finally they get q• wZcos(wt+€-¥)= v• sinwt, from this equation they concluded that v•= q•wZ! This is where I am confused, how can one conclude this according to maths rule. If A*B= C*D, how can we say A=B? This is where I find it odd.

And yes I know the maximum values of the quanties on the sides must be equal as the maximum values of both sin and cos are one(1) but these the two(sin and cos) may not keep their maximum value at same time and hence to take both as one(1) at a single time is to distort this very fact that they may have different values also.

Thanks a bunch!!!

2. Aug 8, 2016

### Let'sthink

There are some conceptual problems here. First - Phasor diagram is no solution it is an equivalent way of understanding the solution. Second The real solution of that differential equation is a complex function and sine and cosine are just real and imaginary parts of that solution. Because as we know the real parts and imaginary parts of a complex number add separately when you add two complex numbers, we find that these real functions are also the solutions of the given equation. Because voltages and currents and charges are real quantities we deal with these real functions either sine or cosine not both. dq/dt is i and i*Z = V.
Third - note that the instantaneous voltages just add. Current is the same for all the components.