Solveeit Logo

Question

Physics Question on Electromagnetic induction

An inductor 20mH20\, mH, a capacitor 100μF100\, \mu F and a resistor 50Ω50 \, \Omega are connected in series across a source of emf, V=10sin314tV\, = \,10 \,sin\, 314\, t. The power loss in the circuit is

A

1.13 W

B

0.79 W

C

2.74 W

D

0.43 W

Answer

0.79 W

Explanation

Solution

Pav=(VRMSZ)2RP_{av} = \left(\frac{V_{RMS}}{Z}\right)^{2} R
Z=R2+(ωL1ωC)2=56ΩZ = \sqrt{R^{2} + \left(\omega L - \frac{1}{\omega C}\right)^{2}} = 56\, \Omega
Pav=(10(2)56)2×50=0.79W\therefore P_{av} = \left(\frac{10}{\left(\sqrt{2}\right)56}\right)^{2} \times50 = 0.79 \,W