{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

W09 314 HW05 solution

# W09 314 HW05 solution - W09 EECS 314 Homework 5 Solution...

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

Solution by Idong Ebong W09 EECS 314 Homework 5 Solution Problem 1 Part 1 Part 2: V S =15V; C 1 =C 2 = 2mF; L 1 =L 2 =1mH; R 1 =R 3 =4Ω ; R 2 =R 5 =1Ω ; R 4 =R 6 Capacitor Voltages =2Ω V C1 = V R2 = R 2 R 1 + R 2 V S = 1 Ω 4 Ω + 1 Ω 15V = 3V V C2 = V S = 15V I L1 = I R4 = V S R 4 = 15 V 2 Ω = 7.5 A I L2 = I R6 = V S R 6 = 15 V 2 Ω = 7.5 A Inductor Currents Component C C 1 Component 2 L L 1 2 Voltage 3V 15V Current 7.5A 7.5A Part 3: E C1 = 1 2 C 1 V C1 2 = 1 2 2 × 10 3 𝐹𝐹 ∗ (3 𝑉𝑉 ) 2 = 0.009 𝐽𝐽 E C2 = 1 2 C 2 V C2 2 = 1 2 2 × 10 3 𝐹𝐹 ∗ (15 𝑉𝑉 ) 2 = 0.225 𝐽𝐽 E L1 = 1 2 L 1 I L1 2 = 1 2 1 × 10 3 𝐻𝐻 ∗ (7.5 𝐴𝐴 ) 2 = 0.0281 𝐽𝐽 E L2 = 1 2 L 2 I L2 2 = 1 2 1 × 10 3 𝐻𝐻 ∗ (7.5 𝐴𝐴 ) 2 = 0.0281 𝐽𝐽 Component C C 1 L 2 L 1 2 Energy in Joules 0.009 0.225 0.0281 0.0281

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

View Full Document
EECS 314 Winter 2009 Homework set 5 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 2 solution Part 1: Calculate Cab in nF: Since there is no connection between node d and any other point in the circuit, no current will flow through it and thus the 50nF capacitor can be neglected in the calculation of Cab. The 60nF and 40 nF caps are in series so the effective capacitance from those two is 60*40/(60+40) = 2400/100 = 24nF. The 24nF effective capacitor is in parallel with the 20nF capacitor and thus they add. The effective capacitance of that configuration is 20nF + 24nF = 44nF.
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