Electrical Engineering - RL Circuits - Discussion

Discussion :: 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

[A]. 5.51
[B]. 55.15
[C]. 90
[D]. 200

Answer: Option B

Explanation:

No answer description available for this question.

Vikas Kumar said: (Jun 3, 2012)  
(140+j0)*(0+j60)/140+j0+0+j60=140angle0*60angle90/140+j60
=8400angle90/152.31angle25
=55.15angle65 so answer is 55.15

Poorni said: (Aug 5, 2016)  
Please explain this.

Muhammad Asif Latif said: (Aug 12, 2016)  
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.

Abdallah Labib said: (Feb 27, 2017)  
Z = R * XL/√(R2 * XL2).
Z= 140 * 60/√(140 * 140 * 60 * 60).
Z= 55.15Ω.

Option B.

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