Maximum Power Transfer Theorem: Proof and Formula

Ans: Applying Thevenin's theorem to the preceding circuit,

Here, 

Thevenin’s resistance (R_th) = (40/3)Ω

Thevenin’s Voltage (V_th) = (200/3) 

Substitute the Thevenin's equivalent circuit for the fraction of the circuit that is the left-side of terminals A and B of the given circuit. The secondary circuit diagram follows.

Using the following formula, we can calculate the maximum power that will be delivered to the load resistor, RL.

P_L, _Max = 2V_Th / 4 R_Th

Substitute 

V_Th = (200/3)V and R_Th = (40/3)Ω in the above formula.

P_L, _Max = (200/3)2/ 4(40/3) = 250/3 watts

Therefore, 

As a result, the maximum power delivered to the load resistor R_L of the given circuit is 250/3 W.

Ans: Let R be replaced first, and V­_o.c be the open circuit voltage (figure 2).

Here,

I = \(\frac{4V}{[(5+1)][2+1] \Omega } = \frac{4}{\frac{5}{2}} = \frac{8}{5}A\)

∴ I₂ = I\(\frac{2}{2+5+1} = \frac{8}{5} \times \frac{1}{4} = \frac{2}{5}A\)

The drop across a-b branch is then

V _a-b = \(\frac{2}{5} \times 1 = \frac{2}{5} V\)

Obviously,

V _o.c = V _a-b + 6V = \(\frac{2}{5} + 6 = \frac{32}{5}V\)

V _o.c = 6.4 V

or,

With reference to Figure 3, calculate the internal resistance of the circuit across x-y.

R_Th = 1||2 + 5|| 1 = \(\frac{17}{3}\)|| 1 = \(\frac{17}{20}\)Ω

= 0.85 Ω

As per the maximum power transfer theorem,

R = R_Th = 0.85 Ω

and P max (max.power) = \(\frac{V^2_{o.c}}{4R} = \frac{6.4^2}{4 \times 0.85}\) ≈ 12W

Ans: Let us first change the "I" source to a "V" source and remove R from the x-y terminals, with the voltage at these terminals equals to Vo.c.

With reference to Figure 5,

\(i = \frac{24}{15}\) = 1.6 A

∴ V _a-b = drop across 5Ω = 1.6 x 5 = 8V

Thus in the left loop,

 – 10 + v_o.c +8 = 0 

or, v_o.c = 2V

R_Th (internal resistance of the circuit as seen through x-y) is calculated once more using Figure 6.

R_Th = \(\frac{10 \times 5}{10 +5}\) + 2 =  5.33 Ω

As per the maximum power transfer theorem,

R = R_TH = 5.33 Ω

and P_max = \(\frac{V^2_{o.c}}{4R}\) = \(\frac{2^2}{4 \times 5.33}\) = 188mV

Ans: R is removed by an open circuit. With reference to figure 8,

\(I = \frac{50}{15} = 3.333 \Omega\)

∴ V _o.c = V _10Ω = 3.333 x 10 = 33.33 V

[It should be noted that b will have a positive polarity as current flows from b to a through 10Ω.]

∴ V _x-y = – 33.33 V = V _o.c

[y terminal being +ve]

With reference to figure 9,

R_Th = \(\frac{5 \times 10}{5+10} + 5\) = 8.33Ω

As per maximum power transfer theorem,

R = R_TH = 8.33 Ω

P_max = \(\frac{V^2_{o.c}}{4R}\) = \(\frac{(-33.33)^2}{4 \times 8.33}\) = 33.34 W

Ans: Let r_L be the resistance connected across x-y to maximize power transfer from source to load. According to the maximum power transfer theorem, r_L should equal the internal resistance of the network when viewed through x-y. Let's call this R_int.

To find R_int,

Here,

R_int = [(1||10) + 2||3] + 5

= \(\frac{(\frac{1 \times 10}{1+10} +2)3}{(\frac{1 \times 10}{1+10} +2)+3}\) + 5 = \(\frac{421}{65} \Omega\)

Thus, the load resistance (r_L) must be having a value \(\frac{421}{65} \Omega\) such that maximum power transfer is possible.

Next, in Figure 10, calculate the open circuit voltage across x-y. Referring to Figure 12, at node 1, KCL provides

i₁ + i₂ + i₃ = 5

or, \(\frac{v - 15}{1} + \frac{v}{5} + \frac{v}{10} = 5\)

[assuming the voltage at node (1) to be v]

or, v + 0.2v + 0.1v = 5 + 10

or, 1.3v = 20 i.e., v = 15.4v

∴ i₂ = \(\frac{v}{5} = \frac{15.4}{5} = 3.1 A\)

This gives

V _o.c = i₂ x 3 = 3.1 x 3 = 9.3 V

The maximum amount of power transfer is given by

P_max = \(\frac{V^2_{o.c}}{4R_{Th}} = \frac{(4.3)^2}{4 \times \frac{421}{65}}\) = 3.34W

Ans: Let us first convert the current source to a voltage source, as shown in Figure 14. R is also replaced by an open circuit. The open circuit voltage at the x-y output terminal is denoted by V_o.c.

The use of the mesh equation in the left loop results in

– 20 + 6 + I (10 + 5 + 2) = 0

or,

\(I = \frac{14}{17}A\)

This gives

V_o.c = 10 + \(\frac{14}{17} \times 2 = 11.65 V\)

Deactivating all sources yields the internal resistance of the circuit when viewed from the x-y terminals, as shown in figure 15.

Here,

R_Th = \(\frac{15 \times 2}{15+7} = \frac{30}{17} = 1.765 \Omega\)

According to the maximum power transfer theorem,

R = 1.765Ω

and Pmax (amount of maximum power transfer)

= \(\frac{V_{TH}^2}{4R} = \frac{11.65 ^2}{4 \times 1.765} = 19.22 W\)

Ans: Figure 17 shows the results of replacing R with an open circuit and designating the nodal voltages as V_x and V_y.

5 = \(\frac{V_x}{5} + \frac{V_x}{2} or \frac{7}{10} x = 5\)

or, V_x = \(\frac{50}{7} = 7.143 V\)

and, V_y = – V_ab = – 4 x 2 = – 8V

In loop a-b-c-d,

– V_x + V_o.c + V_y = 0

or, V_o.c = V_x – V_y = 7.143 – (-8) = 15.143 V

With reference to figure 18,

R_Th = \(\frac{2 \times5}{2+5} + 4 = \frac{10}{7} +4 = \frac{38}{7} = 5.43 \Omega\)

From maximum power transfer theorem,

R = R_Th \(5.43 \Omega\)

and P _max = \(\frac{V^2_{o.c}}{4R_{Th}} = \frac{(15.143)^2}{4 \times 5.43} = 10.96W\)

Ans: With reference to Figure 20, replace R and name the loop currents.

At loop-1,

– 20 + I₁ (20 + 2) – 2I₂ = 0

or, 22I₁ – 2I₂ = 20

or, 11I₁ – I₂ = 10…(1)

At loop-2

(10 + 5 + 2) I₂ – 2I₁ = 0

or, 17I₂ – 2I₁ = 0

or, I₁ = \(\frac{17}{2}I_2 = 8.5 I_2\)   ….(2)

Using (2) in (1),

11(8.5I₂) – I₂ = 10 or 92.5I₂ = 100

ie., I₂ = \(\frac{10}{92.5}\) = 0.108 A

Thus, drop across 5Ω,

i.e., V_5Ω = 0.108 x 5 = 0.54 V

In loop-3, we find that

– 50 – 0.54 + V_o.c = 0

or, V_o.c = 50.54 V

To look for internal resistance across x-y.

R_Th = [(20||2) + 10||5]Ω

= \(\frac{(\frac{20 \times2}{20+2}+10)+5}{\frac{20 \times 2}{20+2}+10+5}\)

= \(\frac{11.82 \times 5}{16.82}\) ≈ 3.5Ω

As per the maximum power theorem,

R = R_Th = 3.5Ω

P_max = \(\frac{V_{o.c}^2}{4R} = \frac{(50.54)^2}{4 \times 3.5} = 182.44W\)