# Stuck with resonance circuit, frustrated

Discussion in 'Homework Help' started by anon17, Mar 22, 2011.

1. ### anon17 Thread Starter New Member

Mar 22, 2011
1
0
Hello! I need a little bit of help for my homework

Here is the problem

An impedance oil takes 144 watts at a lagging factor of 0.6. What values of capacitance and resistance should be connected in series with the coil if the power input to the latter is to remain unchanged and the overall circuit is to be unity? Voltage is 60 v and frequency is 60Hz.

I keep on getting 60Ω for the resistance and 66.31x10μf for the capacitance which are hugely different from the correct answers which are 55.3μf for the capacitance and 24Ω for the resistance.

I need help with the formulas because I am in no doubt that I am using incorrect formulas. I've searched, googled and tried to use lots of them but still I am not able to get the correct answers, I've been trying to solve this for 3 hours now and I am really getting frustrated.

Please lighten my burden.﻿ Thank you.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Yeah - racked my brain to get the values you supplied as correct answers.

Attached is my analysis. Hopefully others will check this as well.

I assumed the 60V was the source rms value.

In summary I have

Rcoil=9Ω, Lcoil=31.83mH.

Hence required C for series resonance (at 60Hz) =221uF and extra R value for maintaining total power of 144W is 16Ω - giving Rtot=25Ω.

The 25Ω value is 'obvious'.

At resonance the effective series resistance for 144W with a 60V rms source = (Vrms)^2/144=3600/144=25Ω.

• ###### Coil Analysis.png
File size:
69.2 KB
Views:
16
Last edited: Mar 23, 2011