{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW26Sol - Platt David Homework 26 Due Nov 9 2005 4:00 am...

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

Platt, David – Homework 26 – Due: Nov 9 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 3) 10 points An e±ective AC voltage oF 145 V at 60 Hz is applied to an RLC circuit with 27 . 4 mH inductor, 0 . 52 μ ² capacitor, and 45 Ω resistor in series. 27 . 4 mH 45 Ω 145 V S 0 . 52 μ ² I What is the e±ective current? Correct answer: 0 . 0284817 A. Explanation: Let : V = 145 V , f = 60 Hz , C = 0 . 52 μ ² = 5 . 2 × 10 - 7 ² , L = 27 . 4 mH = 0 . 0274 H , and R = 45 Ω . ²or an RLC series circuit at f = 60 Hz, X L = 2 π f L = 2 π (60 Hz)(0 . 0274 H) = 10 . 3296 Ω , and X C = 1 2 π f C = 1 2 π (60 Hz)(5 . 2 × 10 - 7 ²) = 5101 . 12 Ω , so the total impedance is Z = q R 2 + ( X L - X C ) 2 = (45 Ω) 2 + h (10 . 3296 Ω) - (5101 . 12 Ω) i 2 ¾ 1 / 2 = 5090 . 99 Ω . Thus the e±ective current is I = V Z = 145 V 5090 . 99 Ω = 0 . 0284817 A . 002 (part 2 oF 3) 10 points What is the power Factor oF the circuit? Correct answer: 0 . 00883915 .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

HW26Sol - Platt David Homework 26 Due Nov 9 2005 4:00 am...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online