Question

One 60V,100W bulb is to be connected to 100V,50 Hzac - source.The potential drop across the inductor is

Answer 1

Answer:

given =  

volt = V₁= 60V         V₂ = 100V

power= P ₁= 100W

f = 50 Hsec.

to find =  the poteintial drop =  ? Vp

solution =

  Vrms =   $\sqrt{60^{2}+Vp{2} }$

∴ 100²  =   60² + Vp²

∴ Vp²  =     10,000 - 36000

∴ Vp² =     6400

∴ Vp  =    80volt  

∴ so the potential drop across the inductor is 80 volt.