## mehbeh 3 years ago For the circuit shown in the figure (ε = 7.00 V, R = 8.00 ), calculate the following quantities.

1. mehbeh

2. Tweedle_Dee

I don't understand what you need calculated?

3. gdmellott

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 <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*(i1-i2 = 0 -7V + 2R*(-i1+i2) + 8R*i2 = 0 or -12V + 4R*i1 + 2R*(i1-i2) = -7V + 2R*(-i1+i2) + 8R*i2 (since 2R*(i1-i2) = 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

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.