Alternating Current Overview

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New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

τ = RC = 10µS
For 0

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

Q = (1/R)√ (L/C) = (1/100)√ (80e-3/2e-6) = 2

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

In damped oscillation

v 2 = 1 8 * 1 T = 1 8 * 10 16 1.6 = 7.8 * 10 14

m a + b v + k x = 0

In the circuit

m d 2 x d t 2 2 + b d x d t + k x = 0

- i R - L d i d t - q C = 0

Comparing equation (i) and (ii)

L d 2 q d t 2 + R d q d t + 1 c q = 0

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2 months ago

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New answer posted

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Vishal Baghel

Contributor-Level 10

ω = 2πf = 1/√ (LC)
L = 1/ (4π²f²C) = 1/ (4π² * 60² * 0.1*10? ) = 70.3 mH

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

L = 0.07H and R = 12Ω
Vs = 220V, 50Hz
XL = 2πfL = 2 (22/7) (50) (0.07) = 22Ω
Z = √ (R² + XL²) = √ (12² + 22²) = √ (144+484) = √628 ≈ 25Ω
i = V/Z = 220/25 = 8.8A
tan φ = XL/R = 22/12 = 11/6
φ = tan? ¹ (11/6)

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2 months ago

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Vishal Baghel

Contributor-Level 10

Z = R 2 + ( X L X C ) 2

Z = ( 1 2 0 ) 2 + ( 1 0 1 0 0 ) 2 = 1 5 0 Ω

ω = 1 L C = 1 1 0 1 * 1 0 4 = 1 0 5

? ω = 2 π f

f = 1 0 3 2 π 1 0 = 5 0 H z

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

ω 0 = 1 0 5 r a d / s e c

P = 16w, 120v at resonance

P = v 2 R 1 6 = ( 1 2 0 ) 2 R R = 1 4 4 0 0 1 6 = 9 0 0 Ω

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

2 π f 1 L C L = λ 2 4 π 2 C c 2 = 9 6 0 2 4 * 3 . 1 4 2 * 2 . 5 6 * 1 0 6 * 9 * 1 0 1 6 =10-7 = 10 * 10-8 H

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