Q7.19  Suppose the circuit in Exercise 7.18 has a resistance of 15\Omega. Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Answers (1)
P Pankaj Sanodiya

The inductance of the inductor L=80mH=80*10^3H

The capacitance of the capacitor C=60\mu F

The resistance of a 15\Omegaresistor 

Voltage supply V = 230V

Frequency of voltage supply f=50Hz

As we know,

Impedance

 Z=\sqrt{R^2+\left ( wL-\frac{1}{wC} \right )^2}

Z=\sqrt{15^2+\left ( 2\pi*50*80*10^{-3}-\frac{1}{2\pi 50*60*10^{-6}} \right )^2}=31.728

Current flowing in the circuit :

I=\frac{V}{Z}=\frac{230}{31.72}=7.25A

Now,

Average power transferred to the resistor:

P_{resistor}=I^2R=(7.25)^2*15=788.44W

Average power transferred to the inductor = 0

Average power transferred to the capacitor = 0:

Total power absorbed by circuit :

P_{resistor}+p_{inductor}+P_{capacitor}=788.44+0+0=788.44W

Hence circuit absorbs 788.44W.

Exams
Articles
Questions