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For the circuit shown in the figure (ε = 7.00 V, R = 8.00 ), calculate the following quantities.

MIT 6.002 Circuits and Electronics, Spring 2007
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I don't understand what you need calculated?
After generating a mesh formula by supposing two current loops. The sources are negative when they generate a negative polarity to the common IR drop. -12V + 4R*i1 + 2R*(i1-i2) = 0 -7V + 2R*(-i1+i2) + 8R*i2 = 0 Note that the steps start near the bottom of this listing. Best love your Algebra. '------------------ Verify: (using the other equation) -7V + 2R*(-201/84A) + 2R*(33/28A) + 8R*(33/28A) = 0 -7V - 201/42V + 33/14V + 66/7V = 0 -588/84V -402/84V + 198/84V + 792/84V = 0 -98V - 67V + 33V + 132V = 0 -165 + 165V = 0

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Redraw the circuit so it is clear:|dw:1330642520865:dw| Simplify using thevenins theorem. Mentally remove the resistor between A & B and short the two sources. Looking into A B you now have two resistors parallel, the 8 Ohm and 4 Ohm for an effective resistance of 2 2/3 Ohms. Now consider the two sources providing and equivalent voltage of 19 volts across 12 Ohms producing a current of 19/12 amps the voltage at A B=12-6 1/3=5 2/3 volts. What we have now is:|dw:1330643773079:dw|
Reverse the polarity of the 12 v source in the upper drawing.

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