Electronics - Inductors - Discussion

Discussion Forum : Inductors - General Questions (Q.No. 32)
32.
A 24 V peak source is connected across a 900 mu.gifH inductor. If the frequency of the source is 22 kHz, what is the reactive power?
2.3 VAR
4.6 VAR
9.3 VAR
0 VAR
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
7 comments Page 1 of 1.

Syed Shakeeb said:   4 years ago
Thanks @Mitesh.

Deepak said:   7 years ago
Reactive Power Q= (E^2)/XL.

where XL =2*π*f*l ==> 2*(3.14)*(22000)*(900*(10^(-6))) ==>124.407 ohm.
E=Vp*0.707 = 24*0.707 = 16.698 volts,

Therefor,
Q = ((16.698)^2)/(124.407),
Q=2.314 VAR,
Reactive power approx 2.3 VAR as per option A.

Nisar said:   10 years ago
@Clara you missed conversion of Vm to Vrms.

MITESH said:   1 decade ago
XL=2*PI*F*L
XL=2*3.14*22K*900u
XL=124.4
P=V^2/R
P=24^2/124.4
P=4.63VAR
BUT HERE 24V PEAK VOLTAGE SOURCE SO
P reactive power=P/2=4.63/2
P=2.3VAR

Rashmi Byakod said:   1 decade ago
Q= Erms^2/XL.

Erms=24/square root of 2 ==> 16.97.

where XL =2*Pi*f*l ==> 2*3.14*22K*900U ==>124.4.

Q = (16.97)^2/(124.4).

Q = 2.3VAR.

Manasa said:   1 decade ago
Given peak or max value..
Since this is ac ckt, rms must be considered..
Erms=24/square root of 2
R=0,implies power factor=0,implies phase angle phi=90 degree
Reactive power=E.I.sin phi
E,I are rms values
I=E/(2*3.14*22k*900u)

Clara said:   1 decade ago
Reactive Power Q= E^2/XL
where XL =2*Pi*f*l ==> 2*3.14*22K*900U ==>124.4
Q = (24)^2/(124.4)
Q=4.63 VAR

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