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5 Answers
- billrussell42Lv 71 month agoFavorite 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 51 month ago
how about reading your electrical chart and work it out yourself , its under ohms law , basic eletrical formula ..
- oldschoolLv 71 month 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 71 month 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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