Physics Alternating Current

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New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Key is open.

i r m s = 1 5 6 0 = 1 4 A

2 0 = 1 4 * 1 0 0 * L L = 0 . 8 H

1 0 = 1 4 . 1 1 0 0 C C = 2 . 5 * 1 0 4 F = 2 5 0 μ F

New answer posted

a month ago

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P
Payal Gupta

Contributor-Level 10

When capacitor is removed

tan 45° = XLXRωL=R

When inductor is removed.

tan45°=XCXR1ωC=R

i0=V0Z=V0R2+ (ωL1ωC)2=V0R=220110=2A

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Capacitive Reactance X C = 1 ω C = 1 2 π f C = 1 2 * 3 . 1 4 * 5 0 * 1 0 * 1 0 6

= 1 0 0 0 3 . 1 4

Vrms=210 V

i rms = V rms X C = 2 1 0 X C

 Peak current =2ims=2*2101000*3.14=0.932

?0.93 A

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

According to transformer ratio,

V S V P = N S N P = 2 : 1

New answer posted

a month ago

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P
Payal Gupta

Contributor-Level 10

At very high frequencies

XC=1ωC0

XL=ωL

Thus equivalent circuit

Z=1+2+2=5Ω

l=2205=44A

New answer posted

a month ago

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P
Payal Gupta

Contributor-Level 10

 P1=cos? =Rz=RXL2+R2

P1=cos? =RR2=12

So,  P1P2=1

New question posted

a month ago

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New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

For capacitor : current leads emf

by 90°.

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

C = ε 0 A d X C = 1 C ω = d ε 0 A ω l 0 = V 0 X C = V 0 ε 0 A ω d = 2 π f V 0 ε 0 A d

l 0 = 2 * 3 . 1 4 * 5 0 * 2 0 * 8 . 8 5 * 1 0 1 2 * 1 2 * 1 0 3 = 2 7 . 7 9 * 1 0 6 A = 2 7 . 7 9 μ A              

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Power dissipated in cycle = P = V 0 l 0 2 c o s ? = V 0 2 2 z * R z = V 0 2 R 2 z  

For resonance, z = R, so

P = V 0 2 2 R = 2 5 0 2 * 2 5 0 2 2 * 5 = 1 2 5 0 0 W a t t = 1 2 5 * 1 0 2 W a t t        

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