Engineering Applying linearity in a circuit

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The discussion focuses on calculating the current Io in a circuit using linearity principles and KCL/KVL techniques. The user successfully calculated several voltages and currents but struggled with finding I3, which is crucial for determining IS. They initially assumed the voltage drop across two resistors equaled the source voltage, leading to an incorrect current calculation of 3mA. A response suggested avoiding rounding errors by using fractions instead of decimals, which helped the user realize their correct value for I3 was actually 3.05mA. The conversation highlights the importance of precision in circuit analysis to meet tolerance thresholds.
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Homework Statement


Find Io in the network in the figure below using linearity and the assumption that Io = 1 mA.
Figure:http://i.imgur.com/Xtu0VmG.jpg

Homework Equations


KCL, KVL, basic analysis techniques.

The Attempt at a Solution


The following values I have calculated correctly:
VR1=9V
VR2=61V
VR3=70V
I1=1mA
I2=2mA
VR4=138V
VS=208v
I3=??
IS=??
IoActual=??

So, my trouble lies in finding I3, as once I find I3, I can easily find IS using KCL, though, the current source is doing something to the voltage drop across I3 that I'm not understanding.

How I tried to solve for I3 was the same way I solved for VS. Since the voltage drop across the source is 208v, then I assumed that the voltage drop across the two resistors on the left summed to 208v as well.

So, using Ohms Law: V/R=I, 208v/(42000Ω+26000Ω)=3mA, which is incorrect. Where did I go wrong?

Thank you
 
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Rounding errors probably. Your analysis is fine. Just use fractions instead of calculating decimals and rounding.
 
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vela said:
Rounding errors probably. Your analysis is fine. Just use fractions instead of calculating decimals and rounding.

Thank you very much Vela. You were correct. 3mA was incorrect, but 3.05mA was correct. (My online homework has a +/- 2% tolerance, so 0.5 was enough to fall outside of the threshold!

Thanks for verifying my methodology.
 

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