# Setting i2 0 we let current on the 1 side be i1 the

This preview shows pages 1–3. Sign up to view the full content.

in coil 1. Setting i2 = 0, we let current on the 1-side be i1. The flux linkage to the core is only the portion that link goes through the 15 H inductor. So, lambda_1 = 15*i1. The flux, phi, is just that divided by N1. To get lambda_2, we multiply phi by N2. So, lambda_2 = 15*(N2/N1)*i1 (with i2 = 0).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
The coupling coefficient is: *** Notice, this is a very low coupling coefficient - the leakage is much larger than the magnetizing inductance. b) The loop equations are straightforward from circuit theory: v1 = L1*di1/dt + M*di2/dt = root2 * 100*sin(10t) v2 = L2*di2/dt + M*di1/dt = -i2*(5 ohms) (Notice the negative at the end, i2 is positive INTO the dot as I've defined for this problem) Text 3.14 a) The loop equations are: 100cos(2t) = - (1 H)*di1/dt - (1 ohm)*i1 - (0.5 H)*di2/dt v = (2 ohms)*(-i2) = (1 H)*di2/dt + (0.5 H)*di1/dt Notice strict adherance to sign convention. Also, this is for time t>0+ b) Re-write with phasor arithmetic: 100 = -jwLs*I1 - R*I1 - jwM*i2 0 = jwM*i1 + (jwLs + Rload)*I2 and we have a 2x2 Matrix (this is the second element, Mathcad starts counting at zero)
Text 3.17 In series, the voltage equation would be: Vtotal = L*di/dt + M*di/dt + L*di/dt + M*di/dt Since they are identical (same L, M) and their voltages just add. Since the current is the samein all, it is Vtota = (2L+2M)*di/dt Leq1 = 2L + 2M For the second case, they are still in series but now the direction of one coil is reversed. The current in coil 1 is i, the current in coil 2 is -i. The voltage of coil 2 substracts from coil 1 this time. Vtotal = L*di1/dt + M*d(i2)/dt - (L*(i2) + M*di/dt) = L*di/dt - M*di/dt + L*di/dt - M*di/dt = (2L - 2M)*di/dt Leq2 = 2L - 2M Solving these two equations, two unknowns: First, just add the equations: Leq1 + Leq2 = 4L Then, substitute: 2M = Leq1 -2L
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern