Series and parrarel circuts combination a. Find the currents I1, I2, I3, I4, I5, and I6. a 5 k R1 = 1 k R7 = 2 k I1 I428 V 6 k R3 = 3 k I3 b c 6 k R2 = 1 k I2 I5 I6 d R4 R5 R6

Question

Jermaine

Engineering

10 June, 20:19

Series and parrarel circuts combination a. Find the currents I1, I2, I3, I4, I5, and I6. a 5 k R1 = 1 k R7 = 2 k I1 I428 V 6 k R3 = 3 k I3 b c 6 k R2 = 1 k I2 I5 I6 d R4 R5 R6

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Kristen Tapia · Answer 1

R1 = 1 k; R2 = 1k; R3 = 3k; R4 = 6k; R5 = 5k; R6 = 6k and R7 = 2k

Vt = 428V

The series and parallel circuit combination is as follow:

(R6║R7 + R5) + R4 + R3║R2 + R1

(6*2/6 + 2) + 5 = 13 / k

(13/2*6/13/6 + 6) = 78/31k

78/3 + 3 = 171/3 = 57k

57k║R2 = 57k║1k = 57/58k + R1 = (57/58 + 1) k = 115/58k = 2k

It = Vt/2k = 428/2000 = 0.2A

∴ I1 = 0.2A

I1 = I2 + I3

Using current divider rules to obtain I2 and I3

∴ I2 = I1 X (I2/I2 + I3) = 0.2 X (1/4) = 0.05A

and I3 = I1 X (I3/I2 + I3) = 0.2 X (3/4) = 0.15A

I3 = I4 + I5, using current divider

I4 = I3 X (I4/I4 + I5) = 0.15 X (6/6 + 5) = 0.08

I5 = 0.15 - 0.08 = 0.07A

I5 = I6 + I7, using current divider

I6 = I5 X (I6/I6 + I7) = 0.07 X (6/6 + 2) = 0.05A

I7 = 0.07 - 0.05 = 0.02A