Question

A circuit containing a $80mH$ inductor and a$60\mu F$ capacitor in series is connected to a $230V,50Hz$ supply. The resistance of the circuit is $15\Omega .$ Obtain the average power transferred to each element of the circuit, and the total power absorbed

Answer 1

$L=80mHC=60\mu F$

Step 1: Find $RMS$ current

Given, $L=80mH=80\times {10}^{-3}H$

$C=60\mu F=60\times {10}^{-6}F$

$R=15\Omega$

Impedance of $LCR$ circuit:

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

Where,
$X_L=2\pi \times 50\times 80\times {10}^{-3}=25.13\Omega$

And

$X_C=\frac{1}{2\pi \times 50\times 60\times {10}^{-6}}=53.05\Omega$

Putting values,

$Z=\sqrt{(15{)}^2+(25.13-53.05{)}^2}$

$Z=31.67\Omega$
Now $RMS$ current is given by,

$I_{rms}=\frac{V_{rms}}{Z}=\frac{230}{31.67}=7.26A$

Step 2: Find average power absorbed by resistor

Then average power absorbed by resistor is,

$P=V_{rms}{I}_{rms}\cos \varphi$

Where,

$\cos \varphi =0^{\circ }\text{and}{V}_{rms}=I_{rms}R$

We get,
$P=(7.26{)}^2\times 15=790.6W$

$P\approx 791W$

Step 3: Find average power absorbed by inductor and capacitor

Since, power absorbed is given by

$P=V_{rms}{I}_{rms}\cos \varphi$

As we know, for inductor and capacitor

$\varphi ={90}^{\circ }$

Hence, $\cos \varphi =0$

Therfore,

Average power to $L$

=Average power to$C=0$

Step 4: Find the total power absorbed by the circuit

Total Power absorbed by circuit=
Average power absorbed by resistor =
$791W$

Final Answer:

Average power to $R\text{is}791W$

Average power to $L=$ Average power to

$C=0$

Total power absorbed$=791W$