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Use conservation of energy to solve for [itex]V_0[/itex].

[tex]P_1\,=\,(30V)\,(-6A)\,=\,-180W[/tex]

[tex]P_2\,=\,(12V)\,(6A)\,=\,72W[/tex]

[tex]P_3\,=\,(V_0)\,(3A)\,=\,(3V_0)W[/tex]

[tex]P_4\,=\,(28V)\,(2A)\,=\,56W[/tex]

[tex]P_5\,=\,(28V)\,(1A)\,=\,28W[/tex]

[tex]P_6\,=\,(10V)\,(-3A)\,=\,-30W[/tex]

[tex]\sum\,p\,=\,p_1\,+\,p_2\,+\,p_3\,+\,p_4\,+\,p_5\,+\,p_6[/tex]

[tex]\sum\,p\,=\,(-180W)\,+\,(72W)\,+\,[(3V_0)W]\,+\,(56W)\,+\,(28W)\,+\,(-30W)[/tex]

[tex]3V_0\,=\,54W[/tex]

[tex]V_0\,=\,18\,V[/tex]

Does this look right?

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# Find V_0 in the circuit (6 elements)

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