A coil of 0.01 H inductance and 1 ω resistance is connected to 200 V, 50 Hz AC supply. Find the impedance of the circuit and time lag between maximum alternating voltage and current.

3 Views|Posted 8 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Payal Gupta | Contributor-Level 10

8 months ago

This is a short answer type question as classified in NCERT Exemplar

L= 0.01H

R= 1ohm

Voltage =200V

Frequency = 50Hz

Z= $\sqrt[]{R^{2} + X_{L}^{2}}$ = $\sqrt[]{1^{2} + ({2 * 3.14 * 50 * 0.01)}^{2}}$ = $\sqrt[]{10.86}$ =3.3ohm

tan $\emptyset = \frac{w L}{R}$ = $\frac{2 π f L}{R} = \frac{2 * 3.14 * 50 * 0.01}{1}$ = 3.14

$\emptyset$ = tan-13.14=72

$\emptyset$ =72 $\frac{π}{180} r a d$

$? t$ = $\emptyset$ /w= $\frac{72 π}{180 * 2 π * 50}$ = $\frac{1}{250}$ s

Upvote

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

1.2K

Exams

6.9L

Reviews

1.8M

Answers

Learn more about...

Physics Ncert Solutions Class 12th 2023

View Exam Details

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering