Question

A $20\text{mH}$ inductor, a $100\mu \text{F}$ capacitor and a $50\Omega$ resistor are connected in series across a source, whose voltage is given by, $V=10\sin \left(314t\right)$. The active power of the circuit is -

• A. $0.8\text{W}$
• B. $0.2\text{W}$
• C. $0.4\text{W}$
• D. $0.6\text{W}$

Answer 1

The correct option is A $0.8\text{W}$

Here,

$X_L=\omega L=\left(314\right)(20\times {10}^{-3})=6.28\Omega$

$X_C=\frac{1}{\omega C}=\frac{1}{314\times 100\times {10}^{-6}}=31.85\Omega$

$R=50\Omega$

So,

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

$\Rightarrow Z=\sqrt{{50}^2+(31.85-6.28{)}^2}\approx 56\Omega$

Now, active power of the circuit,

$P=E_{\text{rms}}{i}_{\text{rms}}\cos \varphi$

$\Rightarrow P=E_{\text{rms}}\left(\frac{E_{\text{rms}}}{Z}\right)\left(\frac{R}{Z}\right)=\frac{(E_{\text{rms}}{)}^{2}R}{Z^2}$

$\Rightarrow P=\frac{(E_0{)}^{2}R}{2Z^2}=\frac{{10}^2\times 50}{2\times {56}^2}\approx 0.8\text{W}$

Hence, option $\left(A\right)$ is the correct answer.