1. ## Cramer's Rule.

Well, it would have been nice if I'd have been taught Cramer's Rule, but our lecturer obviously doesn't work that way.

So I've gone away and looked it up, and it seems relatively straight forward with nice values such as x=2 y=1 z=3 from the samples I've seen...

Why don't my equations look like that?!

$\displaystyle V1(20+j48)-j8V2=j4000<0°$

$\displaystyle j20V1-V2(-200-j60)=-j2000<90°$

Pesky j (i) numbers. '<' denotes polar angle.

I think I'll have to break it down so that like terms read straight downwards, so in this case, there will be a 20V1 in equation 1, but there will effectively be 0V1 in equation 2. Is this thinking correct?

Any help will be much appreciated. Thank you in advance.

2. Originally Posted by AngusBurger
Well, it would have been nice if I'd have been taught Cramer's Rule, but our lecturer obviously doesn't work that way.

So I've gone away and looked it up, and it seems relatively straight forward with nice values such as x=2 y=1 z=3 from the samples I've seen...

Why don't my equations look like that?!
They do look like that. "20+ j48" is every bit as good a number as 2 or 1!

$\displaystyle V1(20+j48)-j8V2=j4000<0°$

$\displaystyle j20V1-V2(-200-j60)=-j2000<90°$

Pesky j (i) numbers. '<' denotes polar angle.
Are you sure about that? You can write a complex number as, say, "$\displaystyle r<\theta$" meaning $\displaystyle r(cos(\theta)+ j sin(\theta))$ but it would be strange to put that "j" in front. "4000<0" is just the real number 4000 and "2000<90" is the imaginary number 2000j. Perhaps they mean for you to multiply them by j and -j? That would give 4000j and +2000.

In any case, Cramer's rule says that the solutions to ax+ by= e and cx+ dy= f are given by
$\displaystyle x= \frac{\left|\begin{array}{cc}e & b \\ f & d\end{array}\right|}{\left|\begin{array}{cc}a & b \\ c & d\end{array}\right|}$

$\displaystyle y=\frac{\left|\begin{array}{cc}a & f \\ c & e\end{array}\right|}{\left|\begin{array}{cc}a & b \\ c & d\end{array}\right|}$

So, here,
$\displaystyle x= \frac{\left|\begin{array}{cc} 4000j & -8j \\ 2000 &-200- 60j \end{array}\right|}{\left|\begin{array}{cc}20+ 48j & -8j \\ 20j & 200+ 60j\end{array}\right|}$

$\displaystyle x= \frac{\left|\begin{array}{cc} 20+ 48j & 4000j \\ 20j & 2000 \end{array}\right|}{\left|\begin{array}{cc}20+ 48j & -8j \\ 20j & 200+ 60j\end{array}\right|}$

The rest is arithmetic.
[quote]I think I'll have to break it down so that like terms read straight downwards, so in this case, there will be a 20V1 in equation 1, but there will effectively be 0V1 in equation 2. Is this thinking correct?[quote]
No, the coefficient of V1 in equation 2 is 20j. You cannot do the real and imaginary parts separately.

Any help will be much appreciated. Thank you in advance.