# Basic electric Circuit with inductors 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?

• No coupling, right? Nov 21, 2013 at 18:19
• Ideal Transformer, so yeah no coupling Nov 21, 2013 at 18:19
• 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 Nov 21, 2013 at 18:35
• 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)) Nov 21, 2013 at 18:37
• @HL-SDK Just to clarify, I am still getting the "Wrong" answer. Is my answer right or is the books answer wrong?? Nov 21, 2013 at 18:51

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!

• 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! Nov 21, 2013 at 19:25
• @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. Nov 21, 2013 at 19:27
• Sue the ba****** I say! Nov 21, 2013 at 19:30
• hahhahahahahaha!! Nov 21, 2013 at 20:07