Question

An inductor 20 mH, a capacitor 100 uF anda
resistor 50 S2 are connected in series across
cource of emf. V = 10 sin314t. The power loss in​

Answer 1

Answer:

The answer will be 0.79 W

Explanation:

According to the problem , the inductance of the inductor is 20 mH

the capacitance of the capacitor is 100 uF and the resistance of 50 S2 of resistor is connected in series.

We know that power loss, p = 1/2 v0 x I0 x cosØ

where  cosØ is the power angle

Now if we consider a plane with R in horizontal axis and (xl-xc) is in vertical axis the the Z is making angle of Ø with the horizontal axis.

Now the value of resistance is 50 S2

Therefore, xl = ωl

         As given V = 10 sin 314t is the emf where ω= 314

now putting the values,xl = ωl= 314 x 20/1000 = 6.28 S2

Now xc = 1/ωc = 1/314 x 100 x 10^(-6) = 31.85 S2

Now z^2 = R^2 + (xl-xc)^2

               = (50)^2 +(25.5)^2 = 3150.25

Therefore, power loss =  p = 1/2 v0 x I0 x cosØ

     = 1/2 v0 x v0/z x R/z = (v0)^2 x R/2z^2  = 100 x 50 /  3150.25 x 2= 0.79 W