Hi Can anyone help me with this Q.

A Resistor & a Capacitor are connected in series across a 150V a.c supply. When the frequency is 40Hz the current is 5A, & when the Frequency is 50Hz the current is 6A. Find the resistance and capacitance of the resistor and capacitor repectively??

I do not know where to start with this one, I am doing a distance learning course so it is hard to get help..

Thanks for any help...

 

With the magnitude of the current you can find the total impedance of the circuit at each frequency.

Then use (R^2)+(Xc^2)=Z^2

to get two equations with two unknowns.

 

Thanks, I did get the total impedances I just didnt know what to do with with them, does this question need a simultaneous equation calculation? Thanks for the help, I thought I knew this area quite well but this one has got me!!

 

Hi Mik3, I am usure where to go from the impedances are you able to help any further?

Thanks
Jimmy

 

You have two equations and two unknowns, thus you need to make simultaneous equations calculation.

 

I'm still not able to get a simultaneous equation from there,

R and X have no coefficients is it still possible??

 

I'm still not able to get a simultaneous equation from there,

R and X have no coefficients is it still possible??

Start with .....

\(\frac{150}{\sqrt{R^{2}+X_{1}^{2}}}=5\)

\(\frac{150}{\sqrt{R^{2}+X_{2}^{2}}}=6\)

where

\(X_{1}=\frac{1}{\omega_{1}C}\)

\(X_{2}=\frac{1}{\omega_{2}C}\)

\(\omega_{1}=2\pi f_{1}\)

\(\omega_{2}=2\pi f_{2}\)

\(f_{1}=40 Hz\)

\(f_{2}=50 Hz\)

 

I'm sorry TNK I still dont see it... is this going to be substitution??

 

I'm sorry TNK I still dont see it... is this going to be substitution??

Yes, you can use substitution or any other method for solving simultaneous equations. If there are no coefficients then the coefficient is one.

 

I cant substitute it too many squar roots and squars for me mik3, just cant get my head round this one

 

\({R}^{2}+\frac{1}{2*pi*40*C}={30}^{2}\) (1) for 40 Hz

\({R}^{2}+\frac{1}{2*pi*50*C}={25}^{2}\) (2) for 50 Hz

solve (1) for \({R}^{2}\)

\({R}^{2}={30}^{2}-\frac{1}{2*pi*40*C}\) (3)

sub (3) into (2)

\({30}^{2}-\frac{1}{2*pi*40*C}+\frac{1}{2*pi*50*C}={25}^{2}\)

Then find C. I will let you find it.

 

Not sure how to set out the calculations on here but is the answer

C= 28.3 micro F??

But also from your equations 1 & 2 shouldnt the Xc part of the equation be squared??

 

mik3 does that sound about right??

 

The correct answer in not 28.3mF it is R=11.6Ω; C=144uF.

 

I know, Xc has to be less than Z,
are you able to show me the workings?

 

Here is a detailed solution for C ....

\(X_{1}^{2}+R^{2}=(\frac{150}{5})^{2}=30^{2}=900 \ \ (1)\\
X_{2}^{2}+R^{2}=(\frac{150}{6})^{2}=25^{2}=625 \ \ (2) \\
\)

Subtract eqn (2) from eqn (1)

\(X_{1}^{2}-X_{2}^{2}=900-625=275\)

\((\frac{1}{\omega_{1}C})^{2}-(\frac{1}{\omega_{2}C})^{2}=275\)

\((\frac{1}{\omega_{1}})^{2}-(\frac{1}{\omega_{2}})^{2}=275C^{2}\)

\(\frac{1}{(2\pi f_{1})^{2}}-\frac{1}{(2\pi f_{2})^2}=275C^{2}\)

\(\frac{1}{251.33^{2}}-\frac{1}{314.16^{2}}=275C^{2}\)

swap sides

\(275C^{2}=1.5831E^{-5}-1.01323E^{-5}=5.7E^{-6}\)

\(C^{2}=\frac{5.7E^{-6}}{275}=2.0727E^{-8}\)

\(C=\sqrt{2.0727E^{-8}}=1.44E^{-4}=144 \mu F\)

 

Thanks tnk, easy when you know how, dont think I would have ever worked that one out though.. do you know of any other good text books to buy, I have 'Hughs electrical & electronic, tenth edition book' that is basicaly what i'm learning from... thanks J

 

Thanks tnk, easy when you know how, dont think I would have ever worked that one out though.. do you know of any other good text books to buy, I have 'Hughs electrical & electronic, tenth edition book' that is basicaly what i'm learning from... thanks J

This is really applied maths - perhaps you just need to practice solving equations of various types.

 

Yes I know what you mean, but I really learn from examples which doesnt always help!

 

Hi tnk, are you able to show the working of the substitution method as mik3 started to show.. I can only get so far but am having trouble isolating 'C'

Thanks for your time J..