- #1

agata78

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## Homework Statement

Determine currents Ia, Ib & Ic in the network?

## Homework Equations

__FIRST EMF__RT = R1 + (R2 x R3) / (R2 + R3)

RT = 47 + (j100 x ( -j75)) / (j100 + (-j75))

RT = 47+ 300j (ohms)

Ohms Law then states that:

I1 = E1 / RT

I1 = 12∠0º / 83-84.48j

I1 = (996+1013.76j) / (14025.87)

Using the Current Division Laws leaves:

I2 = I1 x ( R3 ) / (R2 + R3)

I2 = ( ((996+ 1013.76j) / 14025.87) x ( -j75 )) / (j100 + -j75)

I2 = ( (2988-3041.28j) / 14025.87) ∠0º

I3 = I1 x ( R2 ) / (R2 + R3)

I3 = I1 x ( j100 ) / (j100 + -j75)

I3 = ((3984+4055.04j)/ 14025) ∠0º

__SECOND EMF__RT = R3 + (R1 x R2) / (R1 + R2)

RT = - j75 + ( 47 x j100 ) / ( 47 + j100 )

RT = ? Ω

Ohms Law then states that:

I4 = E2 / RT

I4 = 10∠0º / ?

I4 = ?

Using the Current Division Laws leaves:

I5 = I4 x ( R2 ) / (R1 + R2)

I5 = ? x ( j100 ) / ( 47 + j100 )

I5 = ? x ?

I5 = ? Ω

I6 = I4 - I5

I6 = ? _ ?

I6 = ? Ω

__So:__IA = I1 - I5

IA = ? - ?

IA = ?∠ ?º Ω

IB = I3 - I4

IB = ? - ?

IB = ?∠ ?º Ω

IC = I2 + I6

IC = ? + ?

IC = ?∠ ?º Ω

Can someone please let me know if the information provided is answering the question correctly? Thanks!

In the diagram R3 is j75. However, it is suppose to read -j75.

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