(19) The shown circuit for a wire of resistance 50 2, a coil of induction of resistance 40 N. As the potential difference at the terminals of the coil = the potential difference at the terminals of the wit with a source of frequency 50HZ. Calculate: a. The coefficient of self-induction. 402 b. The current intensity as the e.m.f of the source is v90 volt. c. Using a d.c source providing a similar current instead. d. How to provide a phase angle = zero. (0.095H, 0.1A)

(19) The shown circuit for a wire of resistance 50 2, a coil of induction of resistance 40 N. As the potential difference at the terminals of the coil = the potential difference at the terminals of the wit with a source of frequency 50HZ. Calculate: a. The coefficient of self-induction. 402 b. The current intensity as the e.m.f of the source is v90 volt. c. Using a d.c source providing a similar current instead. d. How to provide a phase angle = zero. (0.095H, 0.1A)

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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