### 5 Answers

- billrussell42Lv 72 months agoFavourite answer
a and b are the same with different values. c requires a Y to delta transformation

b) 300 and 150 in series = 450

60 and 90 in series = 150

now you have 450 and 150 in parallel which is

450•150/600 = 112.5 Ω

that is in series with 20 for a total R of 132.5 Ω

battery current = 8/132.5 = 0.0604 amps

voltage across the parallel pair = .0604 x 112.5 = 6.79 volts

that causes current in both the 300 and 150 of

6.79 / 450 = 0.0151 amps

and it causes current in both the 90 and 90 of6.79 / 150 = 0.0453 amps

and they add up to 0.0604 amps as a check

power in the 150 = I²R = 0.0151²150 = 0.0342 watts

the second I'll leave to another answer.

hint: apply a Y to delta transformation to the 4,5,6 Ω resistors, then you have simple parallel series combos.

- garryLv 62 months ago
how about reading your electrical chart and work it out yourself , its under ohms law , basic eletrical formula ..

- oldschoolLv 72 months ago
Assume i1, i2 an i3 clockwise currents in the left, bottom and top loops. Currents I1, I2, I3, I4 and I5 will be found using those.

The equations are:

-5i1 + 3i2 + 2i3 = -6.6

+3i1 - 12i2 + 5i3 = 0

+2i1 + 5i2 - 13i3 = 0

With the equations you can use any method for simultaneous solutions.

Multiply 3rd eq. by -1.5 and add to 2nd eq,

-39i2 + 49i3 = 0 so i3 = 39i2/49

Multiply the 2nd eq by 5/3 and add to 1st eq.

-17i2 + 31i3/3 = -6.6

i3 = 51i2/31 - 19.8/31 = 39i2/49

i2(51/31 - 39/49) = 19.8/31

i2 = 0.7521

i3 = 0.5986

i1= (3*0.7521+2*0.5986+6.6)/5 = 2.01

I1 = 2.01 - 0.7521 = 1.259

I2 = 0.7521

I3 = i2 - i3 = 0.1535

I4 = I3 + I1 = 1.412

I5 = 2.01 - 1.412 = 0.599

- oubaasLv 72 months ago
fig. (a)

Req = (16 // 6) + 3 = 96/22 + 3 = 162/22 = 81/11 ohm

I = V/Req = 20*11/81 = 220/81 = 2.716 A

fig. (b)

Req = (150 // 450) + 20 = 50*(27/12) + 20 = (1350+240)/12 = 1590/12 ohm

I = V/Req = 8*12/1590 = 96/1590 A

I(300) = I*150/600 = 24/1590 A

I(90) = I*450/600 = 72/1590 A

P(150) = 150*24^2/1590^2 = 34.2 mwatt

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