Derive the condition for maximum power transfer to load for DC circuit

Derive the equation, the condition for the maximum power transmission.T = 3.Tc [7]Were,T - Maximum-tension,Tc - Centrifugal tension

Calculate the maximum power delivered to the load resistor (RL) for the circuit shown in Fig. 5

A V = 37.0 V power supply fully charges a capacitor with capacitance C = 1.03 µF. The capacitor is then connected to a L = 86.7 mH inductor. Find the maximum current (in A) in the resulting oscillations.

.........is the greatest of all demands which have occurred during the specified period of time.(1 Point)total connected loadmaximum demandutilization factormaximum efficiency

To solve this problem, we need to determine the value of the filter inductance \( L \) for a Buck converter that will ensure continuous conduction mode (CCM) at one-third of the maximum output power. Given: - Input voltage, \( V_{in} = 20 \, \text{V} \) - Output voltage, \( V_o = 12 \, \text{V} \) - Maximum output power, \( P_o = 72 \, \text{W} \) - Switching frequency, \( f_s = 400 \, \text{kHz} \) First, calculate the maximum output current \( I_o \): \[ I_o = \frac{P_o}{V_o} = \frac{72 \, \text{W}}{12 \, \text{V}} = 6 \, \text{A} \] At one-third of the maximum output power, the output current \( I_{o, \text{min}} \) is: \[ I_{o, \text{min}} = \frac{I_o}{3} = \frac{6 \, \text{A}}{3} = 2 \, \text{A} \] The duty cycle \( D \) for the Buck converter is given by: \[ D = \frac{V_o}{V_{in}} = \frac{12 \, \text{V}}{20 \, \text{V}} = 0.6 \] The inductor current ripple \( \Delta I_L \) is given by: \[ \Delta I_L = \frac{V_{in} - V_o}{L} \cdot D \cdot \frac{1}{f_s} \] To ensure CCM, the peak-to-peak inductor current ripple \( \Delta I_L \) should be less than twice the minimum output current \( I_{o, \text{min}} \): \[ \Delta I_L \leq 2 \cdot I_{o, \text{min}} \] \[ \Delta I_L \leq 2 \cdot 2 \, \text{A} = 4 \, \text{A} \] Rearranging the inductor current ripple equation to solve for \( L \): \[ L = \frac{(V_{in} - V_o) \cdot D}{\Delta I_L \cdot f_s} \] Substitute the values: \[ L = \frac{(20 \, \text{V} - 12 \, \text{V}) \cdot 0.6}{4 \, \text{A} \cdot 400 \times 10^3 \, \text{Hz}} \] \[ L = \frac{8 \, \text{V} \cdot 0.6}{4 \, \text{A} \cdot 400 \times 10^3 \, \text{Hz}} \] \[ L = \frac{4.8 \, \text{V}}{1.6 \times 10^6 \, \text{A} \cdot \text{Hz}} \] \[ L = 3 \times 10^{-6} \, \text{H} \] \[ L = 3 \, \mu\text{H} \] Therefore, the value of the filter inductance \( L \) should be \( 3 \, \mu\text{H} \) to ensure the converter remains in CCM at one-third of the maximum output power.