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.

Amritesh Gupta said: (Aug 31, 2020)  
The answer is 8.74mA.

Javed said: (Dec 28, 2020)  
R=3.3 = 3300 OHMS.
L=120mH = 0.12H.
f = 2khz = 2000hz.
Source = 12 volt.

XL=2*fl.
2*3.14*2000*0.12/100.
=1507.1ohms.

Z= √R^2+X^2L.
√3300^2+1507.1^2.
√10890000+2271049.
√13161049.
z = 3627.
Z = R * Xc/Z.
z = 3300*1507/3627=1371.1ohms.
Z = V/I or I = V/R.
12/1371=0.00874.
OR 8.75mA.

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