An inductor of 0.1 Henry and a box are connected to AC supply of 25 volts and ω = 100 rad/sec over all power factor of circuit is 1 and the reading of AC ammeter (I rms ) is 5A. Find the power factor of box.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(0.447)
z = x + y 
+ ω L 
= x + 
(y + ω L) for power factor to be
one y + ω L = 0 ⇒ y = – 10
I = 
, x = 
= 
= 5
Impedance of box = 5 – 10 
cos φ = 
= 
= 0.447
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