# Reading of ammeter when power factor is given

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

But could not understand how to use it here ?

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 .

answered Mar 30, 2017 by (2,320 points)
selected Mar 30, 2017 by koolman
The answer is given as D) 3A
Oh no , I did mistake in taking the trigonometric function.
I think Z' should be equal to $\sqrt{R^2 + (L\omega )^2}$
And $\tan \theta = 4/3$
yes , that was an error , thanks I updated the answer ;)