A circuit is constructed with six resistors and two ideal batteries as shown. Th

Question

A circuit is constructed with six resistors and two ideal batteries as shown. The battery emfs are 1 = 18 V and 2 = 12 V. The values for the resistors are: R1 = R5 = 55 , R2 = R6 = 110 , R3 = 44 , and R4 = 117 . The chosen positive directions for the currents are indicated by the directions of the arrows. Figure 1:

(a) Find all currents. Are the assumed current directions correct? (10pt) (b) Find the power dissipated in each resistor. (6pt) (c) Find the power delivered to or by each battery and indicate “delivered to” or “delivered by” for each. Which battery (if any) is being charged? (2pt) (d) Find the total power dissipated and the total power delivered. Are they the same? (2pt)

Please help! I am having a lot of trouble with this problem! Thank you

Explanation / Answer

Applyin Kirchoff's current junction law at all nodes we get following equations

i4 = i5 + i1

i2 = i3 + i4

Current in branch ab ( i.e. through R6 is i5 + i1 = i4)

Applying Kirchoff's voltage law in the left loop

i5R1 -  E1 - i1R3 = 0

55i5 - 18 - 44 i1 =0

55i5 - 44 i1 = 18 ......................(1)

Applying Kirchoff's voltage law in the middle loop

i1R3 + E1 + i4R6 - E2 + i4R2 = 0

putting values we get

44i1 + 220 i4 = -6 ....................(2)

Applying Kirchoff's voltage law in the right loop

i3 R5 + i3 R4 - E2 = 0

putting values

i3 = E2 /( R5 + R4) = 0.07 A

Putting i4 = i1 + i5 in equation 2 and solving equation 1 and 2 we get

i1 = -0.17 A

i5 = 0.19 A

i4 = i1 + i5 = -.17+.19 = 0.02 A

i2 = i3 + i4 = 0.07 + .02 = 0.09 A

Since i1 is negative so the direction i1 is opposite rest all currents have the same direction as shown.

(b) Power dissipated

P = I2 R

P1 = i52 R1 = 1.99 W

P2 = i42 R2 = 0.04 W

P3 = i12 R3 = 1.27 W

P4 = i32 R4 = 0.58 W

P5 = i52 R5 = 1.99 W

P6 = i42 R6 = 0.04 W

(c) If the current direction is coming out of the positive terminal of the battery power is delivered by the battery. We see that both currents (i1 = -0.17 A for E1) and (i2 = 0.09 A for E2) are coming out so power is deleiverd by both batteries

(d) Total power dissipated = P1 +P2 + P3 + P4 +P5 +P6 = 6.31 W

Total power dissipated in the resistors is given by the batteries

Total power dissipated in resistors = Total power delivered by the batteries

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