Homework15 - johnson(rj6247 hw 14 Opyrchal(121014 This...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
johnson (rj6247) – hw 14 – Opyrchal – (121014) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points In a certain series RLC circuit, the rms cur- rent is 12 . 7 A , the rms voltage is 244 V and the current leads the voltage by 37 . What is the total resistance of the circuit? Correct answer: 15 . 3439 Ω. Explanation: Let : I rms = 12 . 7 A , V rms = 244 V , and φ = - 37 . The average power delivered by the generator is dissipated as heat in the resistor: P av = I rms V rms cos φ = (12 . 7 A) (244 V) cos( - 37 ) = 2474 . 81 W , and the power dissipated in the resistor is P av = I 2 rms R , so the resistance is R = P av I 2 rms = 2474 . 81 W (12 . 7 A) 2 = 15 . 3439 Ω . 002 (part 2 of 2) 10.0 points Calculate the reactance X L - X C of the cir- cuit. Correct answer: - 11 . 5624 Ω. Explanation: For the reactance, tan φ = X L - X C R X L - X C = R tan φ = (15 . 3439 Ω) tan( - 37 ) = - 11 . 5624 Ω . 003 (part 1 of 2) 10.0 points In a series RLC circuit, the resistance is 29 Ω, the capacitance is 79 μ F and the inductance is 40 mH. The AC generator provides an rms
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern