A 12 kΩ 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
A. 3.84 ∠20.6° V
B. 12.8 ∠20.6° V
C. 0.3 ∠20.6° V
D. 13.84 ∠69.4° V
Answer: Option A
This Question Belongs to Electrical Engineering >> RL Circuits
Join The Discussion
Comments (2)
Related Questions on RL Circuits
If the frequency is halved and the resistance is doubled, the impedance of a series RL circuit
A. Doubles
B. Halves
C. Remains constant
D. Cannot be determined without values
When the frequency is decreased, the impedance of a parallel RL circuit
A. Increases
B. Decreases
C. Remains constant
D. Is not a factor
R=12000
XL=2pi fL
=2*3.14*8000*90^-3
=4521.6
Z=12000+j4521.6
Z=√12000^2+4521.6^2
=12823.60
O°=tan^-1 4521.6/12000
=20.6
V=12823.6*0.3^-3
=3.84°20.64
Q ka solve ti dete nhi ho