Solved Examples on Maximum Power Transfer Theorem

Demonstrative Video

Problem-1

Determine the maximum power delivered to the resistor $R$ .

Solution-1

\begin{matrix} V_{0} & = \frac{2 K Ω}{5 K Ω} (5) = 2 V V_{o c} & = V_{T H} = V_{40 K Ω} = \frac{40 K Ω}{50 K Ω} (200) = 160 V I_{s c} & = \frac{200}{10 K Ω} = 20 m A R_{t h} & = \frac{V_{t h}}{I_{s c}} = \frac{160 V}{20 m A} = 8 K Ω P_{m a x} & = \frac{V_{t h}^{2}}{4 R_{L}} = \frac{160^{2}}{4 \times 8 K Ω} = 0.8 W \end{matrix}

$\begin{aligned} V_0 & = \dfrac{2 K\Omega}{5 K\Omega}(5) = 2~\mathrm{V} \\ V_{oc} & = V_{TH} = V_{40 K\Omega} = \dfrac{40K\Omega}{50K\Omega}(200)=160~\mathrm{V}\\ I_{sc} & = \dfrac{200}{10K\Omega} = 20~\mathrm{mA} \\ R_{th} & = \dfrac{V_{th}}{I_{sc}} = \dfrac{160\mathrm{V}}{20\mathrm{mA}}=8~K\Omega \\ P_{max} & = \dfrac{V_{th}^2}{4R_L} = \dfrac{160^2}{4\times 8K\Omega} = 0.8~\mathrm{W} \end{aligned}$

Problem-2

Determine the value of $R_L$ for maximum power transfer.

Solution-2

\begin{matrix} V_{1} & = V_{x} + 4 (I_{1} - \frac{V_{x}}{4}) = 4 I_{1} R_{t h} & = \frac{V_{1}}{I_{1}} = 4 Ω \end{matrix}

$\begin{aligned} V_1 & = V_x +4\left(I_1-\dfrac{V_x}{4}\right)=4I_1\\ R_{th} & = \dfrac{V_1}{I_1} = 4~\Omega \end{aligned}$

Problem-3

Find the value of R in the circuit such that maximum power transfer takes place. What is the amount of this power?

Solution-3

$\begin{aligned} \text{Mesh-1}~3i_1-2i_2 & = 4\\ \text{Mesh-2}~8i_2-2i_1 & = 0\\ \text{solving}~i_2 & = \dfrac{2}{5}~\mathrm{A} \\ \because 1\times i_2 + 6 & = V_{oc} \\ \Rightarrow V_{oc} & = \left(6+\dfrac{2}{5}\right)\\ & =\dfrac{32}{5}~\mathrm{V} \end{aligned}$