Answer:
Explanation:
The image that is supposed to be attached to the question is displayed in the diagram below.
Applying Nodal Analysis at node 1;
[tex]\dfrac{V_o -50}{12.5*10^{-3}} + \dfrac{V_o}{50*10^3}+\dfrac{V_o-7500 \ in}{10*10^3}=0[/tex]
where;
[tex]in = \dfrac{V_o}{50*10^3}[/tex] (from the circuit)
= [tex]\dfrac{V_o-50}{12.5}+\dfrac{V_o}{50} + V_o -\dfrac{7500 *V_o }{\frac{50*10^3}{10}}=0[/tex]
= [tex]V_o [ \dfrac{1}{12.5}+\dfrac{1}{50}+\dfrac{1}{10}-\dfrac{75}{500}] = \dfrac{50}{12.5}[/tex]
= [tex]V_o[ \dfrac{500*500+12.5*5000+12.5*5000*5-75*12.5*500}{12.5*50*10*500}]= \dfrac{50}{12.5}[/tex]
= [tex]V_o = 80 \ volts[/tex]
[tex]in = \frac{80}{50*10^3}= 1.6 mA \\ \\ 7500*in = 120 volts \\ \\ I = \frac{120-80}{10(10^3} =4*10^{-3} Amps \\ \\ \\ \\ P_{generated} = 75000*in*I \\ \\ P_{generated} = 120*4*10^{-3} \\ \\ P_{generated} = 480 \ MW[/tex]
