Products
Rewards 
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY 
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY 
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY 
HELP DESK

Need Help? We got you covered.

Chapter 7

Q. 7.6

Finding the Values of L and C for a Series-Loaded Resonant Inverter to Yield Specific Output Power

A series resonance inverter in Figure 7.8a with series loaded delivers a load power of P_L = 1  kW at resonance. The load resistance is R =10  \Omega. The resonant frequency is f_0 = 20  kHz. Determine (a) the dc input voltage V_s, (b) the quality factor Q_s if it is required to reduce the load power to 250 W by frequency control so that u = 0.8, (c) the inductor L, and (d) the capacitor C.

Step-by-Step

Verified Solution

a. Because at resonance u = 1 and |G(j\omega )|_{max}=1, the peak fundamental load voltage is V_p=V_{i(pk)}=4V_s/\pi .

P_L=\frac{V_p^2}{2R} =\frac{4^2V_s^2}{2R\pi ^2}       or              1000=\frac{4^2V_s^2}{2\pi ^2\times 10}

which gives V_s = 110 V.

b. To reduce the load power by (1000/250 = ) 4, the voltage gain must be reduced by 2 at u = 0.8. That is, from Eq. (7.35),

|G(j\omega )|=\frac{1}{[1+Q_s^2(u-1/u)^2]^{1/2}}                                      (7.35)

we get 1 + Q_s^2(u – 1/u)^2 = 2^2, which gives Q_s = 3.85.

c. Q_s is defined by

Q_s=\frac{\omega _0L}{R}

or

3.85=\frac{2\pi \times 20 kHz\times L}{10}

which gives L = 306.37  \mu H.

d. f_0=1/2\pi \sqrt{LC}or 20 kHz -1/[2\pi \sqrt{306.37 \mu H\times C} ] which gives C = 0.2067  μF.