Question

An inductor of inductance $20mH$, a capacitor of capacitance $100\mu F$ and a resistor of resistance $50\Omega$ are connected in series across a source of emf, $V=10\sin \left(314t\right)$. The power loss in the circuit is

• A. $0.79W$
• B. $0.43W$
• C. $2.74W$
• D. $1.13W$

Answer 1

The correct option is A $0.79W$

$V_0=10V,\omega =314rad/sP=V_{rms}{i}_{rms}\cos \varphi =V_{rms}\left(\frac{V_{rms}}{Z}\right)\left(\frac{R}{Z}\right)=\frac{{V_{rms}}^{2}R}{Z^2}{X}_L=\omega L=\left(314\right)(20\times {10}^{–3})=6.280\Omega X_C=\frac{1}{\omega C}=\frac{1}{314\times 100\times {10}^{-6}}=31.84\Omega R=50\Omega Z=\sqrt{{\left(X_C-X_L\right)}^2+R^2}=\sqrt{{\left(31.84-6.28\right)}^2+{50}^2}=56\Omega \Rightarrow P=\frac{{\left(\frac{10}{\sqrt{2}}\right)}^2\times 50}{{56}^2}=0.79W$