Reading of ammeter when power factor is given

Question

I know power factor (cos $\phi$) =(Average power)/(rms power)=R/Z

But could not understand how to use it here ?

Answer 1

Powrer factor of circuit = 1
$R/Z= 1$ so $ R=Z $
$ Z= \sqrt {(wL-1/wC)^2 +R^2}$

but Z=R so,$
R^2=R^2 +(wL-1/wC)^2$
So $wL= 1/wC $
$1/wC= 40$

where $R $ would be resistance due to box only , as we don't see any other resistance
(provided the ammeter was ideal)
So for box,

Let $Z' $ be reactance of box.

So $ R/Z'= $
$ R/wL=3/5==> R/40 =cot\theta=3/4$
this gives R=30

Now R= Z for complete circuit,
so I= 3A ,
I think it would be 3A then .

Comments

The answer is given as D) 3A

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Oh no , I did mistake in taking the trigonometric function.

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I think Z' should be equal to $\sqrt{R^2 + (L\omega )^2}$
And $\tan \theta = 4/3$

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yes , that was an error , thanks I updated the answer ;)