Electrical Engineering - RL Circuits - Discussion
Discussion Forum : RL Circuits - General Questions (Q.No. 11)
11.
A 140 
resistor is in parallel with an inductor having 60 inductive reactance. Both components are across a 12 V ac source. The magnitude of the total impedance is
resistor is in parallel with an inductor having 60 inductive reactance. Both components are across a 12 V ac source. The magnitude of the total impedance is5.51
55.15
90
200
Discussion:
4 comments Page 1 of 1.
Abdallah labib said:
9 years ago
Z = R * XL/√(R2 * XL2).
Z= 140 * 60/√(140 * 140 * 60 * 60).
Z= 55.15Ω.
Option B.
Z= 140 * 60/√(140 * 140 * 60 * 60).
Z= 55.15Ω.
Option B.
Muhammad Asif Latif said:
9 years ago
Impedance Z = 1/Y,
Admittance Y = sqrt ( (1/140) 2 + (1/60) 2)
Y = 0.018.
Z = 1/Y,
Z = 1/0.018,
Z = 55.14.
Admittance Y = sqrt ( (1/140) 2 + (1/60) 2)
Y = 0.018.
Z = 1/Y,
Z = 1/0.018,
Z = 55.14.
Poorni said:
9 years ago
Please explain this.
Vikas kumar said:
1 decade ago
(140+j0)*(0+j60)/140+j0+0+j60=140angle0*60angle90/140+j60
=8400angle90/152.31angle25
=55.15angle65 so answer is 55.15
=8400angle90/152.31angle25
=55.15angle65 so answer is 55.15
(1)
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