Oct 13, 2014 #41 H Hatmpatn Member level 3 Joined Sep 15, 2014 Messages 59 Helped 0 Reputation 0 Reaction score 0 Trophy points 6 Activity points 515 Yes!! I got it! With input in matlab: -((12000*i*R)/(-12000*i - (16+15*i)*R+12000000*C*R)) Where the real part of the equation is: R=1131.36*R+848520000*C*R+848520+1060.65*R And the complex part: C=848520+1060.65*R-1131.36*R-848520000*C*R-12000*R We get R=171.788 Ω and C=8.40447*10^-6F Which gives us when we put it into the equation: 70.71-70.71i
Yes!! I got it! With input in matlab: -((12000*i*R)/(-12000*i - (16+15*i)*R+12000000*C*R)) Where the real part of the equation is: R=1131.36*R+848520000*C*R+848520+1060.65*R And the complex part: C=848520+1060.65*R-1131.36*R-848520000*C*R-12000*R We get R=171.788 Ω and C=8.40447*10^-6F Which gives us when we put it into the equation: 70.71-70.71i
Oct 13, 2014 #42 T The Electrician Full Member level 5 Joined Jul 13, 2010 Messages 300 Helped 141 Reputation 282 Reaction score 143 Trophy points 1,323 Location Pacific NW Activity points 4,607 Congratulations! Here's how I did it in Mathematica:
Oct 13, 2014 #43 H Hatmpatn Member level 3 Joined Sep 15, 2014 Messages 59 Helped 0 Reputation 0 Reaction score 0 Trophy points 6 Activity points 515 I just thought of something.. We want the equation to be =70.71-70.71j, but shouldnt it be =70.71+70.71j? In post #23 there was a positive sign but since that it was negative?
I just thought of something.. We want the equation to be =70.71-70.71j, but shouldnt it be =70.71+70.71j? In post #23 there was a positive sign but since that it was negative?
Oct 13, 2014 #44 T The Electrician Full Member level 5 Joined Jul 13, 2010 Messages 300 Helped 141 Reputation 282 Reaction score 143 Trophy points 1,323 Location Pacific NW Activity points 4,607 Remember the maximum power transfer theorem. The impedance of the load must be the complex conjugate of the source impedance for maximum power transfer.
Remember the maximum power transfer theorem. The impedance of the load must be the complex conjugate of the source impedance for maximum power transfer.
Oct 13, 2014 #45 H Hatmpatn Member level 3 Joined Sep 15, 2014 Messages 59 Helped 0 Reputation 0 Reaction score 0 Trophy points 6 Activity points 515 Yes ofcourse! Well, I'd like to thank you The Electrician for your help, and it was also pleasant to see that tony_lth got engaged!
Yes ofcourse! Well, I'd like to thank you The Electrician for your help, and it was also pleasant to see that tony_lth got engaged!