The series resonant inverter in Figure 2 has the following c

The series resonant inverter in Figure 2 has the following characteristics L_1=L_2 = L = 40 mu H C = 10 mu H R = 1 Ohm V_s = 100 Ohm Output voltage frequency f_0 = 5 KHz Transistor turn -off t_off = 5 mu S

Solution

A.Vpp= I/2fC

where I is current, f is frequency of ac power and C is capicitance

I=Vs/R= 100/1= 100A

Vpp= 100/ 2*5000*10*10^-6

Vpp= 1000 V

B. Output Power (Po)

Po= I² * R

Po= 100² * 1

Po= 10⁴

P_avg = I²_RMS* R

10⁴ = I²_RMS * 1

I_RMS =100 A