# 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

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.