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enter image description here

Its required to find \$v_o\$. The question should be easy but I am not getting the same answer as offered by the textbook which is 2cos(400t-53). I am getting 5cos(400t-53). I tried doing in two ways: 1)reflecting the right impedance to the left and then converting the votlage back to the right circuit, 2) Using KVL and ideal transformers equation. I will show the latter here.

\$\frac{V_1}{V_2}=\frac{N_1}{N_2}=2\$. So, \$V_0=0.5V_1\$ But, we also know that \$I_2=2I_1\$. And that \$V_0=5I_2\$ where \$I_2\$ is the clockwise current in the secondary circuit. This implies that \$V_1=20I_1\$ (Eq 1) Using KVL, in circuit 1: \$-25\angle0 +40jI_1+10I_1+V_1=0\$ (Eq 2) Putting Eq 1 in Eq 2, and calculating \$I_1\$, we can then find \$V_1\$(using eq 1) and then \$V_0\$ which is half of \$V_1\$.

Have I done a mistake or is the answer in the book wrong?

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  • \$\begingroup\$ No coupling, right? \$\endgroup\$
    – Shabab
    Nov 21, 2013 at 18:19
  • \$\begingroup\$ Ideal Transformer, so yeah no coupling \$\endgroup\$
    – user29568
    Nov 21, 2013 at 18:19
  • \$\begingroup\$ I'd reflect the right impedance to the left since you only have to transform one element. I believe you'll still have to use the turns ratio to get the correct v0 though. If you get the same answer with that method, I'd start to doubt the book \$\endgroup\$
    – HL-SDK
    Nov 21, 2013 at 18:35
  • \$\begingroup\$ Oops I misread and mis-wrote. I actually did what you said(reflecting the right impedance to the left) and I got the same answer.(5cos(400t-53)) \$\endgroup\$
    – user29568
    Nov 21, 2013 at 18:37
  • \$\begingroup\$ @HL-SDK Just to clarify, I am still getting the "Wrong" answer. Is my answer right or is the books answer wrong?? \$\endgroup\$
    – user29568
    Nov 21, 2013 at 18:51

1 Answer 1

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Sanity check

The 0.1H impedance at 2\$\pi\$F = 100 rad/sec is 40 ohms.

Reflecting \$R_0\$ to the left gives you 20 ohms and therefore the total circuit impedance is: - \$\sqrt{40^2 + (10 + 20)^2}\$ = 50 ohms.

Given that the drive voltage is 25 V, this has to mean the current from the supply is 0.5 A.

This current flows through the 20 ohms (reflected) producing a voltage of 10V. This has a power of 5W and therefore if the resistor was placed back on the right-hand side you'd have the same power and expect a voltage of: -

V = \$\sqrt{5W\times 5\Omega}\$ = 5V

Looks like we're all getting 5V!

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  • \$\begingroup\$ This book is making me go crazy!!!! This is not the only answer that is not agreeing with me. At least, you brought my sanity back :). I feel a tad more confident now. Thanks! \$\endgroup\$
    – user29568
    Nov 21, 2013 at 19:25
  • \$\begingroup\$ @user29568, typos in textbooks can drive a student and instructor crazy. So, if you suspect a typo, try simulating the circuit with your favorite circuit simulator. \$\endgroup\$
    – Alfred Centauri
    Nov 21, 2013 at 19:27
  • \$\begingroup\$ Sue the ba****** I say! \$\endgroup\$
    – Andy aka
    Nov 21, 2013 at 19:30
  • \$\begingroup\$ hahhahahahahaha!! \$\endgroup\$
    – user29568
    Nov 21, 2013 at 20:07

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