Explanation:
circuit can be considered as thevenin's equivalent
The equivalent circuit is
Given,
[tex]\begin{aligned}&I_{L}=3 m A\\&R_{L}=1.2 \mathrm{K} \Omega\end{aligned}[/tex]
By applying KVL to circuit
[tex]\begin{array}{l}I=\frac{V_{th} }{R+h+R_{L}} \\3 m=\frac{V _{th} }{R_{m}(k \Omega)+1.2 K}\end{array}\\[/tex]
[tex]V_{t h}=3 R+h+3 \times 1.2[/tex]
[tex]I=\frac{Vth}{Rth+R_{L}}[/tex]
[tex]2.5 m=\frac{Vth}{Rth(k \Omega)+2 K}[/tex]
[tex]V_{th}=2.5 \mathrm R_{th} +5[/tex]
By equating, we get
[tex]R_{th}=2 \cdot 8 K \Omega[/tex]
According to maximum power transfer theorem, maximum power in transfered to load
[tex]R_{L}=R_{th}[/tex]
[tex]R_{th}=2 \cdot 8 K \Omega[/tex]
