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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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