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mehbeh
Group Title
For the circuit shown in the figure (ε = 7.00 V, R = 8.00 ), calculate the following quantities.
 2 years ago
 2 years ago
mehbeh Group Title
For the circuit shown in the figure (ε = 7.00 V, R = 8.00 ), calculate the following quantities.
 2 years ago
 2 years ago

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Tweedle_Dee Group TitleBest ResponseYou've already chosen the best response.0
I don't understand what you need calculated?
 2 years ago

gdmellott Group TitleBest ResponseYou've already chosen the best response.0
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*(i1i2) = 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 <Checks okay) ' Step#3: (Repeat with i2's value, usually in the other equation, to find i1's value.) 12V + 4R*i1 + 2R*i1  2R*33/28A = 0 12V + 6R*i1  33/14V = 0 i1 = ( 33/14V + 12V ) / 6R i1 = 201/84A i1 = 2.392857142857142857~ A ' Step#2: Substitute the 'equal' in one equation. 7V + 2R*(5V/8R  3/2*i2) + 2R*i2 + 8R*i2 = 0 7V + 5/4V  3R*i2 + 10R*i2 = 0 7R*i2 = 33/4V i2 = 33/28A i2 = 1.17857~A ' Step#1: Find an i2 'equal' for i1 ( or if possible an i2 result). 12V + 4R*i1 + 2R*(i1i2 = 0 7V + 2R*(i1+i2) + 8R*i2 = 0 or 12V + 4R*i1 + 2R*(i1i2) = 7V + 2R*(i1+i2) + 8R*i2 (since 2R*(i1i2) = 2Ri1  2Ri2 and 2R*(i1+i2) = (2Ri1  2Ri2) then they combine across the equal sign. 5V + 4R*i1 =  4Ri1 + 4Ri2 + 8R*i2 8R*i1 = 5V + 12R*i2 i1 = 5V/8R + 3/2*i2
 2 years ago

radar Group TitleBest ResponseYou've already chosen the best response.0
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=126 1/3=5 2/3 volts. What we have now is:dw:1330643773079:dw
 2 years ago

radar Group TitleBest ResponseYou've already chosen the best response.0
Reverse the polarity of the 12 v source in the upper drawing.
 2 years ago

radar Group TitleBest ResponseYou've already chosen the best response.0
dw:1330651277322:dw
 2 years ago
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