Electrical Engineering - RL Circuits - Discussion

Discussion :: RL Circuits - General Questions (Q.No.17)

17. 

A 3.3 k resistor and a 120 mH coil are in parallel. Both components are across a 2 kHz, 12 V ac source. The total current in the circuit is

[A]. 8.74 mA
[B]. 874 mA
[C]. 874 A
[D]. 8.74 A

Answer: Option A

Explanation:

No answer description available for this question.

Vikas Kumar said: (Jun 3, 2012)  
XL=2*3.14*2*1000*.120=1507.2,R=3300,

z=3300angle0*1507.2angle90/3300+j1507.2

z=4973760angle90/3627angle27

z=1371.31angle63

i=v/z,12/1371.31=8.75 mA

Madan C said: (Sep 4, 2016)  
I think the answer is 3.30MA.

XL = ωL.
Z = sqrt(R^2 + xL^2).
I = v/z.

Vishu said: (Mar 19, 2017)  
Can anyone say what is exact answer for this?

Hemanth Kumar said: (Jul 23, 2017)  
Ir = 12/(3.3)mA = 3.37mA.
IL = -j12/(2*22/7*2K*120m)A = -j7.96mA,
I= Ir+IL.
= 3.37-j7.96mA,
= 8.644angle-67.

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