The V = 10.90 V battery in the figure below is removed from the circuit and rein

Question

The V = 10.90 V battery in the figure below is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. (R = 10.80 ?.)

(a) Find the current in each branch.
upper branch (A) =
middle branch (A) =
lower branch (A) =

(b) Find the potential difference Vab of point a relative to point b. (V)

The V = 10.90 V battery in the figure below is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. (R = 10.80 ?.) (a) Find the current in each branch. upper branch (A) = middle branch (A) = lower branch (A) = (b) Find the potential difference Vab of point a relative to point b. (V)

Explanation / Answer

Let I1=Current from V=10.9 Volts battery

I2=Current from 5 Volts battery

From Kirchoffs Juction rule Current through R is

I3=I1+I2-----------------------1

Applying KVL in outer loop

10.9-(2+3)*I1-10.8*(I1+I2)=0

15.8I1+10.8I2=10.9---------------------2

Applying KVL in lower loop

5+(4+1)*I2+10.8*(I1+I2)=0

10.8I1+15.8I2=-5-----------------------3

solving 2 and 3

I1=1.7 A

I2=--1.48 A

I3=1.7-1.48=0.22 A

b)

Vab=3*1.7-4*(-1.48)=11.02 Volts

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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