Solveeit Logo
Login

Question

Question: An inductor 200mH, capacitor 500\(\mu F\)and resistor 10\(\Omega\)are connected in series with a 100...

An inductor 200mH, capacitor 500μF\mu Fand resistor 10Ω\Omegaare connected in series with a 100V variable frequency ac source. What is the frequency at which the power factor of the circuit is unity?

A

10.22Hz

B

12.4Hz

C

19.2Hz

D

15.9Hz

Answer

15.9Hz

Explanation

Solution

: Here, L=200mH=200×10−3H=0.2HL = 200mH = 200 \times 10^{- 3}H = 0.2H

C=500μF=500×10−6=5×10−4FC = 500\mu F = 500 \times 10^{- 6} = 5 \times 10^{- 4}F

R=10ΩR = 10\Omega and Vrms=100VV_{rms} = 100V

Power factor , =cos⁡φ=RZ=1= \cos\varphi = \frac{R}{Z} = 1 (Given),

∴Z=R\therefore Z = R

R2+(XL−XC)2=R\sqrt{R^{2} + (X_{L} - X_{C})^{2}} = R

R2+(XL−XC)2=R2R^{2} + (X_{L} - X_{C})^{2} = R^{2}

⇒XL−XC=0\Rightarrow X_{L} - X_{C} = 0

or XL=XCX_{L} = X_{C} (This is resonance condition)

2πυL=12πυC2\pi\upsilon L = \frac{1}{2\pi\upsilon C}

4π2υ2LC=14\pi^{2}\upsilon^{2}LC = 1

∴υ=12πLC=12×3.140.2×5×10−4\therefore\upsilon = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2 \times 3.14\sqrt{0.2 \times 5 \times 10^{- 4}}}

=12×3.14×10−2=1006.28=15.9Hz= \frac{1}{2 \times 3.14 \times 10^{- 2}} = \frac{100}{6.28} = 15.9Hz