Electrical Engineering - RL Circuits - Discussion
Discussion Forum : RL Circuits - General Questions (Q.No. 8)
8.
A 470 
resistor and a 200 mH coil are in parallel. Both components are across a 1.5 kHz ac source. The total admittance of the circuit, in polar form, is
resistor and a 200 mH coil are in parallel. Both components are across a 1.5 kHz ac source. The total admittance of the circuit, in polar form, is12.60 
–76° mS
–76° mS12.1876° mS
76° mS218 –76° mS
–76° mS12.18 –76° mS
–76° mSDiscussion:
7 comments Page 1 of 1.
ASHRAF said:
9 years ago
Y = 1/Z.
=1/(470 + j1884.96),
.0005 < -75.99.
=1/(470 + j1884.96),
.0005 < -75.99.
(1)
Joesmarti said:
3 years ago
Y= 1/R + (1/jX).
Y = 1/470 + 1/j*2π*1.5k*200m.
Y = 2.12 - j 0.5351 mS.
Y = 2.19 <-14 or 2.19 <76.
Y = 1/470 + 1/j*2π*1.5k*200m.
Y = 2.12 - j 0.5351 mS.
Y = 2.19 <-14 or 2.19 <76.
(1)
Jansi said:
10 years ago
xL = 2*22/7*F*L.
= 2*22/7*1500*200m.
= 1884.9 OHM.
Y = 1/R + 1/XL.
= 1/470 + 1/1884.9.
= 2.12 m + 0.5 m.
lyl = sqrt (2.12 m^2 + 0.5 m^2).
= 2.17 m.
PHASE ANGLE= tan-1 (XL/R) = tan-1 (1884.9/470).
=75.99 = 76.
= 2*22/7*1500*200m.
= 1884.9 OHM.
Y = 1/R + 1/XL.
= 1/470 + 1/1884.9.
= 2.12 m + 0.5 m.
lyl = sqrt (2.12 m^2 + 0.5 m^2).
= 2.17 m.
PHASE ANGLE= tan-1 (XL/R) = tan-1 (1884.9/470).
=75.99 = 76.
Roderick acuesta said:
9 years ago
The answer is 5.14x10^-4 < -75.999.
Saroj said:
9 years ago
Solve it in another method.
Kesiena said:
6 years ago
I thought the answer is D.
Ajit kumar yadav said:
5 years ago
@Jansi.
Right, thanks.
Right, thanks.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers