Electrical Engineering - RL Circuits - Discussion
Discussion Forum : 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
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 is8.74 mA
874 mA
874 
A
A8.74 A
ADiscussion:
6 comments Page 1 of 1.
Amritesh Gupta said:
5 years ago
The answer is 8.74mA.
(1)
VIKAS KUMAR said:
1 decade ago
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
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:
9 years ago
I think the answer is 3.30MA.
XL = ωL.
Z = sqrt(R^2 + xL^2).
I = v/z.
XL = ωL.
Z = sqrt(R^2 + xL^2).
I = v/z.
Vishu said:
8 years ago
Can anyone say what is exact answer for this?
Hemanth Kumar said:
8 years ago
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.
IL = -j12/(2*22/7*2K*120m)A = -j7.96mA,
I= Ir+IL.
= 3.37-j7.96mA,
= 8.644angle-67.
Javed said:
5 years ago
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.
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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