Alternating Current Overview

Get insights from 72 questions on Alternating Current Overview, answered by students, alumni, and experts. You may also ask and answer any question you like about Alternating Current Overview

Follow Ask Question
72

Questions

0

Discussions

3

Active Users

0

Followers

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

At resonance VL = VC

c o s ? = V R V R 2 + ( V L V C ) 2 = 1

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

X C = 1 ω C

On increasing ω  XC decreases so Z decreases, current in circuit increases.

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

H D . C . H A . C . = I 2 R I r m s 2 R = 1

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

E = V R 2 + ( V L V C ) 2

E = ( 6 0 ) 2 + ( 1 0 0 2 0 ) 2

E = 100V

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

R = 2 Ω

L = 2 mH

E = 9V

i = ε 2 R = 9 v 4 Ω = 2 . 2 5 A

Just after the switch 'S' is closed, the inductor acts as open circuit.

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f = 1 2 π L C = 1 2 π ( 0 . 5 * 1 0 3 ) * ( 2 0 0 * 1 0 6 )

f = 1 0 4 2 π 1 0 5 * 1 0 2 H z [ T a k i n g π = 1 0 ]

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

X? = X? ⇒ ωL = R ⇒ 2 * 3.14 * 50 * L = 20 ⇒ L = 63.7mH

New answer posted

a month ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Diameter = main scale reading + (circular scale reading * least count)
Diameter = 0 + (52 * 0.01 mm) = 0.52 mm = 0.052 cm.
In the RLC circuit:
Given I? = 10√2 A, so I? = I? /√2 = 10 A.
V? = √ [V? ² + (V? - V? )²] = √ [40² + (40 - 10)²] = √ [1600 + 30²] = √ [1600 + 900] = √2500 = 50 V.
Impedance Z = V? / I? = 50 V / 10 A = 5 Ω.
For Hindi: I? = 10√2 A, V? = 50V, Z = V? /I? = 50/ (10√2) = 5/√2 Ω.

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

z = √ [R² + (X? - X? )²] = √ [6² + (4-10)²] = 6√2 Ω

Power factor = cosφ = R/z = 6/ (6√2) = 1/√2

New answer posted

a month ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

I = I? sin (ωt) + I? cos (ωt)

This can be written as I = I? sin (ωt + φ), where I? = √ (I? ² + I? ²)

A hot wire ammeter reads the rms value of the current.

I_rms = I? /√2 = √ (I? ² + I? ²)/√2

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

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 682k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.