NEET | 2018Alternating CurrentEasy

Question

An inductor 20 mH, a capacitor 100 μF and a resistor 50Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is

Options

A.

0.79 W

B.

0.43 W

C.

2.74 W

D.

1.13 W

Solution

V0 = 10V, ω = 314 rad/s
P = Vrms irms cosϕ
=VrmsVrmszRz=Vrms2Rz2XL = ωL = (314) (20 × 10–3) = 6.280Xc=1 ωc=1314×100×10-6=31.84ΩR = 50Ωz=Xc-XL2+R2=31.84-6.282+(50)2=50Ω⇒P=10Ω2×50562=0.79W

Create a free account to view solution

View Solution Free
Topic: Alternating Current·Practice all Alternating Current questions

More Alternating Current Questions