Question

An inductor $200mH,$ capacitor $500\mu F$ and resistor $10\Omega$ are 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 1

Answer: (D)

$15.9Hz$

Here $L=200mH=200\times {10}^{-3}H=0.2H$
$C=500\mu F=500\times {10}^{-6}=5\times {10}^{-4}F$
$R=10\Omega and{V}_{rms}=100V$
Power factor $=cos\varphi =\frac{R}{Z}=1$ (Given)

$\therefore Z=R\Rightarrow \sqrt{R^2+(X_L-X_C{)}^2}=R$

$R^2+(X_L-X_C{)}^2=R^2\Rightarrow X_L-X_C=0$

or $X_L=X_C$ (This is resonance condition)

$2\pi \nu L=\frac{1}{2\pi \nu C}\Rightarrow 4{\pi }^2{\nu }^{2}LC=1$

$\therefore \nu =\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\times 3.14\sqrt{0.2\times 5\times {10}^{-4}}}$

$=\frac{1}{2\times 3.14\times {10}^{-2}}=\frac{100}{6.28}=15.9Hz.$