# 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? – Shabab Nov 21 '13 at 18:19
• Ideal Transformer, so yeah no coupling – user29568 Nov 21 '13 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 – HL-SDK Nov 21 '13 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)) – user29568 Nov 21 '13 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?? – user29568 Nov 21 '13 at 18:51

## 1 Answer

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! – user29568 Nov 21 '13 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. – Alfred Centauri Nov 21 '13 at 19:27
• Sue the ba****** I say! – Andy aka Nov 21 '13 at 19:30
• hahhahahahahaha!! – user29568 Nov 21 '13 at 20:07