Final answer:
In a square-wave inverter with a DC source of 96 V and a series RL load of 5 Ohms and 100 mH, the maximum current is 19.2 A, and the minimum is 0 A. The load current expression involves solving differential equations with transient analysis, while voltage and current Fourier amplitudes require calculations using fundamental frequency and harmonic components.
Step-by-step explanation:
Calculation of Current and Fourier Series Amplitudes
When dealing with a square-wave inverter with a DC source of 96 V and a series RL load with R = 5 Ohm and L = 100 mH, we can calculate the minimum and maximum current as follows:
- The maximum current (I_max) occurs when the voltage is at its peak, which is 96 V. As a square wave alternates between the peak voltage and zero, the peak current is simply the voltage divided by the resistance, I_max = V/R = 96 V / 5 Ω = 19.2 A.
- The minimum current (I_min) is 0 A, occurring when the voltage crosses zero.
The expression for the load current for a square-wave inverter with an RL load can be complex due to the reactive component of the circuit. However, during steady-state conditions, the current through an inductor with a square wave applied can be approximated by its transient responses at each switching interval. This involves solving differential equations and considering the boundary conditions at each switching moment.
For the Fourier series amplitudes of the voltage and current:
- The amplitude of the voltage Fourier series terms for a square wave can be calculated using the formula An = (4V/pi)(1/n) for n being odd integers (1, 3, 5, 7, 9,...). In our case, V = 96 V, hence, the amplitude for the first term (n=1) is A1 = (4 * 96 / pi)(1/1).
- The current Fourier series terms amplitudes can be determined by considering the impedance of the inductor for each harmonic component of the frequency of the applied square wave voltage. This involves a more complex calculation factoring in the inductive reactance XL = 2πfL, with f being the frequency of the nth harmonic.