How Do You Calculate Currents ib and ia in a Circuit with Multiple Resistors?

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To calculate the currents ib and ia in a circuit with multiple resistors, start by analyzing the current Ix, which is 0.4 mA, and how it splits into ib and i1. The equations derived from Ohm's law (V=IR) include relationships between the currents and resistances, specifically R2*ib and R1*(0.4mA - ib). The confusion arises regarding the independent current source and its potential voltage, which can complicate the analysis if not properly accounted for. By calculating the voltage Vx from Ix and the resistor combinations, the values of ia and ib can be determined effectively. Understanding these relationships is crucial for solving the circuit.
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Homework Statement


Nilsson-p3.26.png

Determine ib and ia

Ix = 0.4 mA, R1 = 15k ohm, R2 = 19k ohm, R3 = 23k ohm, R4 = 15k ohm

Homework Equations



V=IR


The Attempt at a Solution


I am assuming there are 4 unknown currents in this diagram.
Ix splits into ib and i1.

0.4mA - ib - i1 = 0
i1 - i2 = ia

R2*ib - Vo + R1*(0.4mA - ib) = 0

R3*ia + R4*(0.4mA - ia) = 0

I am confused about the independent current source. In some problems, I have seen it carry a voltage. So if I include a voltage in my analysis for that loop I will have another unknown.

Vx(0.4mA) + R1(0.4mA - ib) + R3(ia) = 0
 
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Hint: Vx can be calculated from Ix and the series/parallel combination of the resistors. Then Ia and Ib are easily calculated.
 
Thank you, that makes sense
 
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