Answers and Discussions for

ECE Ph.D. Qualifying Exam PE-1 August 2010



Recycled Information from Problem 1

Create and Solve Magnetic Equivalent Circuit

Magnetic Equivalent Circuit of UU Core Inductor

$ \begin{align} G(g) &= \frac{\mu_r w_u d_u}{2 d_s d_u + d_u^2 + w_s w_u + w_u^2 + \mu_r d_u g} \\ \lambda_1(i_1,i_2,g) &= \frac{1}{2}N^2 \mu_0 \ell (i_1 + i_2) G(g) \\ \lambda_2(i_1,i_2,g) &= \lambda_1(i_1,i_2,g) \end{align} $

Problem 2

The magnetic equivalent circuit (MEC) used in Problem 1 provides useful information for Problem 2. The current waveforms are $ i_1(t) = 0 \; \forall t $ (open circuited) and $ i_2(t) = I_2 \cos(\omega_e t) $ while the airgap distance $ g $ remains fixed.

The voltage equation for the first coil may be written down.

$ \begin{align} v_1(t) &= r_1 \cancelto{0}{i_1(t)} + \mathit{p}\lambda_1(t) \\ v_1(t) &= \frac{d}{dt} \frac{1}{2}N^2 \mu_0 \ell (\cancelto{0}{i_1(t)} + i_2(t)) G(g) \\ v_1(t) &= \frac{1}{2}N^2 \mu_0 \ell G(g) \frac{d}{dt} \left[I_2 \cos(\omega_e t)\right] \\ v_1(t) &= -\frac{1}{2}N^2 \mu_0 \ell G(g) I_2 \omega_e \sin(\omega_e t) \end{align} $

The answer to this problem is mostly just substitution and a time derivative of a sinusoid.

$ \begin{equation} \boxed{v_1(t) = -\frac{1}{2} \frac{N^2 \mu_r \mu_0 \ell w_u d_u}{2 d_s d_u + d_u^2 + w_s w_u + w_u^2 + \mu_r d_u g} \omega_e I_2 \sin(\omega_e t)} \end{equation} $


Discussion



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Sean Hu, ECE PhD 2009