1. 04 pointsI Previous Answers HRWS 27 P045 In the figure below, the resistances

Question

1. 04 pointsI Previous Answers HRWS 27 P045 In the figure below, the resistances are R1 " 0.8 o, R.-201, and the ideal batteries have emts Zi . 2.0 v, and t2· . 3.5 n, (a) What is the current through each battery? (Take upward to be positive) battery 1 -576 battery 2 288 battery 3 x A (b) What is the potential difference V-Vs Do you see that there are three loopf to choose from? Did you choose two of them? Did you draw a current for two of the branches? You need to g direction. Did you somehow combine them (they either subtract or add) to symibolze the current in the third branch? As you walked around two of t you write a loop equation of the potential differences you encountered? VERY IMPORTANT: Did you get the signs of those potential differences correc currents, did you walk from point b to point a, keeping track of the potential differences along the way? Read the eBook Section 27-7 Mutloop Cirouits 12. O 1 points 1Previous Answers HRW9 27 P057 Switch S in the figure below is closed at time t " o, to begin charging an initially uncharged capacitor of capacitance C·11.0 throughresstor esc 80

Explanation / Answer

Applying KVL through first loop and second loop,

E1 - i1R1 - (i1 +i2) R2 - E2-i1R1 =0

E3 - i2 R1 - (i1 +i2) R2 - E2-i2R1 =0

Putting the values, 2-0.8i1 - 2(i1+i2) - 3.5- 0.8i1 =0

and 3.5- 0.8i2 - 2*(i1+i2) - 3.5 - 0.8 i2 = 0

Solving the equations i1= - 0.6027 A

and i2 = 0.3348 A

Current i1 = - 0.6027 A answer

B) current = - (- 0.6027+ 0.3348) =0.2679 A

C) current = i2 = 0.3348 A

D) Va-Vb = E2 +(i1+i2) R2 = 3.5 - 0.2679*2 = 2.964 V

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.