Electrical Engineering - RL Circuits - Discussion
Discussion Forum : RL Circuits - General Questions (Q.No. 10)
10.
A 12 k
resistor is in series with a 90 mH coil across an 8 kHz ac source. Current flow in the circuit, expressed in polar form, is I = 0.3
0° mA. The source voltage, expressed in polar form, is
resistor is in series with a 90 mH coil across an 8 kHz ac source. Current flow in the circuit, expressed in polar form, is I = 0.30° mA. The source voltage, expressed in polar form, is3.8420.6° V
20.6° V12.820.6° V
20.6° V0.320.6° V
20.6° V13.8469.4° V
69.4° VDiscussion:
2 comments Page 1 of 1.
Anand said:
1 decade ago
Source voltage = current * impedence.
Impedence = R+jX ( X is reactance of the inductor i.e., X=2*3.14*f*L= 4521.6ohm).
Impedence = 12000+j4521.6.
in polar form Impedence = 12823.60<20.64.
Hence Voltage = 0.3*10^(-3)<0 * 12823.60<20.64 = 3.847<20.646.
A is correct.
Impedence = R+jX ( X is reactance of the inductor i.e., X=2*3.14*f*L= 4521.6ohm).
Impedence = 12000+j4521.6.
in polar form Impedence = 12823.60<20.64.
Hence Voltage = 0.3*10^(-3)<0 * 12823.60<20.64 = 3.847<20.646.
A is correct.
Haji Bilal said:
9 years ago
Can anyone explain this answer clearly?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers