Mumbai University > FE > Sem 1 > Basic Electrical and Electronics Engineering

Marks: 3 M

Year: May 2014

$\because I=\dfrac{V_{TH}}{R_{TH}+R_L} \\ \text{Power delivered to the load=PRL} \\ P_{RL}=I^2R_L \\ P_{RL}=\bigg(\dfrac{V_{TH}}{R_{TH}+R_L}\bigg)^2R_L\\ P_{RL}=\dfrac{V^2 _{TH} R_L}{(R_{TH}+R_L)}..............(1) \\ \text{Condition for maximum power transfer is} \\ \dfrac{dP_{RL}}{dR_L}=0 \\ By (1) \\ \dfrac{d}{dR_L}\bigg[\dfrac{V^2_{TH}R_L}{(R_{TH}+R_L)^2}\bigg]=0 \\ V^2_{TH}\dfrac{d}{dR_L}\bigg[\dfrac{R_L}{(R_{TH}+R_L)^2}\bigg] \\ V^2_{TH}\bigg[\dfrac{(R_{TH}+R_L)^2-(R_L)[2(R_{TH}+R_L)]}{(R_{TH}+R_L)^4}\bigg]=0 \\ (R_{TH}+R_L)^2-2R_L(R_{TH}+R_L)=0 \\ R_{TH}^2+R_L^2+ {2R_{TH}R_L} -2R_{TH}R_L-2R^2_L=0 \\ R_{TH}^2-R^2_L=0 \\ R_{TH}=R_L \\ \text{This is the condition for maximum …