Find the Capacitance when given the Power Factor

[jaus tail]

Homework Statement

w = 1 rad/s
power factor is 0.8 lag

Homework Equations

Using Phasor we can do

The Attempt at a Solution

Since pf = 0.8 lag. I source lags V source by 36.87 degrees.
Vc is voltage across Capacitor = Voltage across ( Resistor + inductor)

Solving this I get C = 0.5
But book says C = 0.125
I had actually solved this sum many weeks ago and got right answer then but I cannot find the book now and am struggling to solve again. Can anybody help me?

 

[cnh1995]

Your phasor diagram is correct, but it is difficult to get anything out of it.

Assume the source to be of 1V.
Draw the phasor parallelogram of currents whose one phasor is the current in the RL branch (known) and other side as the capacitor current (unknown). Then you can solve it using simple geometry. The answer is indeed 0.125F.

 

[jaus tail]

I also get 0.125 if I make this change above.
In circled part, if I take Ic angle as 0 degrees I get answer, but why should Ic angle be zero. Ir = | Ic/(square root 2) / C |at angle -135 degrees as per equation 1. So when I substitute Ir angle 0, shouldn't I substitute Ic/1.414 / C and the angle -135 degrees as well?
But I get right answer when I substitute only magnitude of Ir and not the angle in the KCL equation. Why?

I also get the right answer if I substitute in KCL equation: Ic angle 135 by 1.414 C Ir angle 135.
But shouldn't it be angle 135 + 135? I'm not able to understand why only magnitude should be substituted and not the phase angle from equation 1... to get the right answer?

 

[jaus tail]

Can somebody please explain this to me? Why in the red circled part above has the angle not been substituted? If I replace 135 with zero then I get the answer.

 

[cnh1995]

Sorry for the late reply, I'd almost forgot about this thread.

In your working, why have you multiplied I_R by √2? It seems you forgot to multiply I_c by √2 as well, since √2 should not appear in the equations as it is just a scaling factor and gets canceled out on both the sides.
I haven't checked your solution thoroughly, but this is probably why you are having trouble with phase angles (sin 135°=1/√2 and maybe it cancels the √2 in your voltage/current magnitude).

Anyways, I won't recommend this approach for GATE. I believe the phasor approach would be easier and quicker.
All the best!