Test: Maximum Power Transfer - Electrical Engineering (EE) MCQ

The maximum power transferred is equal to:
I²RL, when load resistance is variable.
Es²/4RL, when load impedance is variable.
In both the cases, the maximum power depends on the load resistance.

The circuit can draw maximum power only when source resistance is equal to the load resistance.

When Source impedance is equal to Z, its load impedance is the complex conjugate of Z which is Z’. Only under this condition maximum power can be drawn from the circuit.

Rs is the real part of the complex number ZL. Hence when we find the complex conjugate the real part remains the same whereas the complex part acquires a negative sign.

 On shorting the voltage sources:
RL=3||2+4||5.

The two nodal equations are:
(VA-10)/3+VA/2=0
(VB-20)/4+VB/3=0
On solving the two equations, we get VA=4V, VB=8.571V.
Vba=VA-VB= 4.57V.

Correct Answer :- a

Explanation : On shorting the voltage sources:

R_L = 3||2+4||3 

= 1.20 + 1.71

= 2.91 ohm.

The two nodal equations are:

(V_A-10)/3 + VA/2 = 0

(V_B-20)/4 + VB/3 = 0

On solving the two equations, we get V_A=4V, V_B=8.571V.

V_AB = V_A - V_B 

= 4V – 8.571V 

= -4.57V.

E_th = 4.57V

The maximum power transferred = E_th2/4RL. Substituting the given values in the formula, we get P_max = 1.79W

Maximum power transfer does not imply maximum efficiency. If the load resistance is smaller than source resistance, power dissipated at the load is reduced while most of the power is dissipated at the source then the efficiency becomes lower.

Efficiency=output/input*100.
Under maximum power conditions:
Output=I²RL; Input=2*I²RL.
Thus efficiency=50%.

 Maximum power transfer to the load is preferred over maximum efficiency in both amplifiers and communication circuits since in both these cases the output voltage is more than the input.