Answers and Discussions for

ECE Ph.D. Qualifying Exam ES-1 August 2009



Recycled Information from Problem 1

Create and Solve Magnetic Equivalent Circuit

Magnetic Equivalent Circuit of Plunger

$ \begin{equation} \lambda_1(i,g) = \frac{2N^2 \mu_0 w t d i}{2 t g + w c} \end{equation} $

Problem 2

The magnetic equivalent circuit (MEC) used in Problem 1 is valid because it is given that the plunger remains walled in by the base on each side. The current waveform is $ i(T) = i \; \forall T $, and the plunger position waveform is $ g(T) = \left.v_g T\right|_{v_g = 1 \, \frac{\mathrm{m}}{\mathrm{s}}} $.

The voltage equation for the coil may be written (winding resistance is negligible).

$ \begin{align} v_1(i, T) &= \cancelto{0}{r_1 i_1} + \mathit{p}\lambda_1 \\ v_1(i, T) &= \frac{d}{dT} \left[\frac{2N^2 \mu_0 w t d i}{2 t g(T) + w c}\right] \\ v_1(i, T) &= \frac{2N^2 \mu_0 w t d \frac{di}{dT} \left(2 t g(T) + w c\right) - 2N^2 \mu_0 w t d i \left(2 t \frac{dg}{dT}\right)}{\left(2 t g(T) + w c\right)^2} \\ v_1(i, T) &= \frac{2N^2 \mu_0 w t d \left[2 t v_g T \frac{di}{dT} + w c \frac{di}{dT} - 2 t v_g i\right]}{\left(2 t v_g T + w c\right)^2} \end{align} $

The answer to this problem hinges upon being able to perform the Quotient Rule correctly on an expression from the start of the previous problem. The voltage equation may be simplified further since $ \frac{di}{dT} = 0 $ (derivative of a constant is zero) and $ v_g = 1 \, \frac{\mathrm{m}}{\mathrm{s}} $.

$ \begin{equation} \boxed{v_1 = \frac{-4N^2 \mu_0 w t^2 d i}{\left(2 t T + w c\right)^2}} \end{equation} $


Discussion



Back to ES-1, August 2009

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett