Question

Average power dissipated in a series AC RLC circuit connected to a source whose voltage is given by, $E=E_0\sin \left(\omega t\right)$, if $X_L=X_C$ is -

$R$ is the resistance of the resistor.

• A. $\frac{(E_0{)}^2}{R}$
• B. $\frac{(E_0{)}^2}{2R}$
• C. $\frac{2(E_0{)}^2}{R}$
• D. $\frac{3(E_0{)}^2}{R}$

Answer 1

The correct option is B $\frac{(E_0{)}^2}{2R}$

Average power dissipated in a series AC RLC circuit,

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

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

$\Rightarrow P=\frac{(E_0{)}^{2}R}{2Z^2}=\frac{(E_0{)}^{2}R}{2(R^2+(X_C-X_L{)}^2)}$

$\Rightarrow P=\frac{(E_0{)}^{2}R}{2R^2}=\frac{(E_0{)}^2}{2R}[\text{Given}{X}_L=X_C]$

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