Maker Pro
Maker Pro

Parallel LRC circuit analysis help

spartan09

Apr 16, 2014
1
Joined
Apr 16, 2014
Messages
1
Hello I have a parallel RLC circuit and I am struggling to find the following parameters:

1) Total Circuit current
2) phase angle relative to the supply
3) Total Circuit Impedance using j operation (complex notation)
4) Power dissipated in the resistor

Link to Circuit diagram:
http://i695.photobucket.com/albums/vv313/patel5557600/kjadsbcksjdb_zps4027ce59.png
II have also attached a simple circuit diagram in this thread

My workings so far:

V=12V peak= 12/√2=8.49 V_RMS
f=10KHz
R=10Ω
L=150μH
C=330nF

1 - Circuit Current

I_R=V/R=8.49/10=0.849A
I_L=V/X_L =8.49/9.42=0.901A
I_C=V/X_L =8.49/48.23=0.176A

I= √(〖I_R〗^2+(I_L - I_C )^2 )
I= √(〖0.849〗^2+(0.901-0.176)^2 )
I= 1.116A

2 - Phase Angle

〖G=〗⁡〖1/R〗
〖G=〗⁡〖1/10〗
〖G=〗⁡〖0.1 s〗
〖θ=tan〗^(-1)⁡〖G/Y〗
〖θ=tan〗^(-1)⁡〖0.1/0.132〗

θ=⁡〖37.15〗

3 - Impedance Using J operator

1/Z=√((1/R)^2+(1/X_L -1/X_C )^2 )
1/Z=√((1/j10)^2+(1/(-j9.42)-1/j48.23)^2 )
1/Z=√(((1∠0)/(10∠90))^2+((1∠0)/(9.42∠-9... )
Y=1/Z=√((0.1∠-90)^2+((0.106∠-90)-(0.020... )
Y=1/Z=√((0.1∠-90)^2+((0.106∠-90)-(0.020... )

4 - Power Dissipated

P=VIcosϑ
P=(8.49)(1.116)cos⁡(37.15)
P=7.56W



Can someone please confirm these results? If I have gone wrong, could someone please correct me. Any help would be much appreciated.

Thank you
 

Attachments

  • kjadsbcksjdb.png
    kjadsbcksjdb.png
    19.2 KB · Views: 127

duke37

Jan 9, 2011
5,364
Joined
Jan 9, 2011
Messages
5,364
This is partly beyond me but 4. is wrong.
There is 8.49V rms across the 10R resistor. The other compnents do not come into it since they do not dissipate power.

Also in 1. Ic is given as a function of Xl. I presume this is a typo.
 

Arouse1973

Adam
Dec 18, 2013
5,178
Joined
Dec 18, 2013
Messages
5,178
Look very good. Only thing I can see that might be wrong is you state R as being j10 as being an imaginary number. Shouldn't it be just 1/10. The average power in the coil and capacitor if ideal components will be zero. The phase angles are +90 and -90 for ideal inductor and capacitor that being the voltage and current. The resistor does not have a phase.
Adam
 

Laplace

Apr 4, 2010
1,252
Joined
Apr 4, 2010
Messages
1,252
My calculation gives the same total current as above.
 

Attachments

  • parallel-RLC.png
    parallel-RLC.png
    17 KB · Views: 130
Top