A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
For the circuit shown in the figure (ε = 7.00 V, R = 8.00 ), calculate the following quantities.
 2 years ago
For the circuit shown in the figure (ε = 7.00 V, R = 8.00 ), calculate the following quantities.

This Question is Closed

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

gdmellott
 2 years ago
Best ResponseYou've already chosen the best response.0After 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

radar
 2 years ago
Best ResponseYou've already chosen the best response.0Redraw 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

radar
 2 years ago
Best ResponseYou've already chosen the best response.0Reverse the polarity of the 12 v source in the upper drawing.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.