Capacitor to be connected to increase power factor

[Nax13]

A resistance 'R' ohm and inductance of 'L' henry are connected across 240V, 50Hz supply. Power distributed in the circuit is 300W, and the voltage across R is 100V. In order to improve the power factor to unity, the capacitor to be connected in series should have a value of __.
a) 43.7 uF b) 4.37 uF c) 437 uF d) 0.437 uF

Step 1) Using P=V/I, got I=1.25A
Step 2) Using R=V_R/I, got R=80 ohm
Step 3) Now since total voltage is 240V, V_L is 240-100=140V
Step 4) Using X_L=V_L/I, got X_L=112 ohm.
Step 5) For power factor to be unity, X_L=X_C, which means that X_C should also be equal to 112 ohm, which leads me to the answer that C is 28.4uF. (since X_C=1/2πfC)

Where did I go wrong? Any help would be appreciated.

 

[uart]

Where did I go wrong?

First line (ignoring the typo. P=V/I), you used P=VI to find I. This is incorrect, VI is the apparent power if you use the total voltage. If however you just use the voltage across the resistor ...

 

[tiny-tim]

Hi Nax13!

First find the https://www.physicsforums.com/library.php?do=view_item&itemid=303" of the RL circuit.

 

[uart]

BTW. It is a poor exercise to refer to this a merely "power factor correction". Series resonating out a reactive element like this does a lot more than merely correcting power factor. In this case for example it increases the total power by a multiple of about 6. If the load was designed to run at 300W then it would likely be destroyed.

This is why PFC is normally done with a parallel capacitor.

 

[Nax13]

Why is it necessary to calculate the impedance?
Btw, are all my other steps right?

 

[uart]

Why is it necessary to calculate the impedance?

Because you want to find the reactance "X". So find the resistance "R", find the impedance magnitude |Z| and use R^2 + X^2 = |Z|^2 to solve for X.

 

[Nax13]

Thanks for your help, uart and tim! :)