Hi,
this is the method: https://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
if I understand correctly the method: We calculate the voltage on AB terminals, because this is the voltage we get when rearranging the circuit with just only one voltage source and in series connected resistor? The same is for the resistor.
How do we calculate if we have inductors and capacitors and not just only resistors? Do we consider impedance, let's say for the inductor = jwL? The problem is that we now have complex numbers. How do we solve it then?
Thanks.
Hi there,
A little prerequisite for AC circuits is the theory of complex numbers. If you have dealt with them before then you will find AC circuits to be exceptionally EASY. Much, much, much easier than more general transient circuits where you can have signals that are other than sinusoidal. You will be very happy to learn about complex numbers. AC circuit theory will look easy after that.
The main difference is not about the individual numbers involved, but how we handle them and how to do the main operations like multiply, divide, add, and subtract. I'll give a quick intro here...
Addition of two complex numbers a+b*j and c+d*j:
(a+b*j)+(c+d*j)=a+c+(b+d)*j
So you see all we do is keep the real and imaginary parts separated.
Subtraction is the same, just subtract, so we end up with:
a-c+(b-d)*j
Multiplication get a little more tricky, but just follows the rules of algebra:
(a+b*j)*(c+d*j)=a*c+c*b*j+a*d*j+b*d*j^2
and then we can note that j^2=-1 because sqrt(-1) is j.
Division is a little harder, but not too bad.
The main idea in any of these is that we have to do more than one operation for a pair of complex numbers rather than just one operation like addition. In the end we get another complex number as the result.
Knowing just how to add, subtract, multiply, and divide complex numbers, can get you pretty far into AC circuit theory. Just add a small table of powers of j and you will find it is not that hard.
Also, there is automatic math software that can get you going faster, and advanced calculators that you can use to make this short work.
Questions welcome, i'll try to get back soon.
For the addition above a quick example is to add 1+2*j plus 3+4*j:
1+3+(2+4)*j=4+6*j
See how easy that was?