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I am working on the following circuit:
enter image description here
Simplifying the circuit:
enter image description here
After simplifying, my answers do not match that of values I get from simulating the circuit in multisim. Here's my calculation: $$Voltage V_B$$ $$V_B=6(\frac{5000}{5000+(32.35714286\times10^3)})$$ $$V_B=0.8030592734V$$ $$6-0.8030592734=5.20V$$
$$Voltage Vc$$ $$V_c=5.2(\frac{(32.35714286\times10^3)}{5000+(32.35714286\times10^3)})$$ $$V_c=4.50V$$
The simulations I did on multisim show that the voltages for Vb and Vc are: $$V_B=5.161V$$ $$V_c=5.035V$$
I don't know if its the simplification I'm doing wrong or my approach towards the voltage divider rule. I need help.

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    \$\begingroup\$ Your first circuit has a resistor between C and ground, but your 2nd does not, so they are certainly not equivalent. \$\endgroup\$ – The Photon May 1 '19 at 15:21
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Your first circuit has a resistor between C and GND. You seem to have added it to your bottom resistor. You also seem to have your 1k||3k resistor calculation incorrect. 3k*1k/3k+1k is 750 ohms. Thus you should end up with a circuit like this:

schematic

simulate this circuit – Schematic created using CircuitLab

From here, it is simpler to do Ohms Law to find the current in the circuit, then find the voltage drops of all the resistors.

If you must use the voltage divider rule, then you need to know if you are finding the voltage drop(s) across the resistor(s), or the voltage at the points with respect to GND, because that will make a difference to how you calculate it.

Of course, you always have the option to simplify to 2 resistors with the R2 component in the voltage divider as (R2+R3) too. It depends what your task is.

Another thing I noticed is you used your answer from Vb as your input voltage for your second divider equation. You should still use 6V as the supply for both equations. If you do that, you'll end up calculating answers that agree with your simulation.

I ended up with Vb = 5.1608V and Vc = 5.035V

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  • \$\begingroup\$ I am asked to calculate the node voltages for B and C using the voltage divider rule \$\endgroup\$ – AugieJavax98 May 1 '19 at 15:44
  • \$\begingroup\$ So that's the node voltages with respect to GND I assume? In that case, it should be fairly straightforward now. You have demonstrated you know the voltage divider rule. Simplify the circuit to how I have it and apply it again, You should get answers that match your simulation, as I did \$\endgroup\$ – MCG May 1 '19 at 15:50
  • \$\begingroup\$ Yes it is with respect to GND. I will try again. \$\endgroup\$ – AugieJavax98 May 1 '19 at 15:55
  • \$\begingroup\$ When I get the answer from the divider equation, I have to subtract it from the source voltage right? \$\endgroup\$ – AugieJavax98 May 1 '19 at 16:00
  • \$\begingroup\$ No, you do the normal voltage divider equation, and you get your answer. You don't have to subtract anything from the source. You should get answers that match your sims \$\endgroup\$ – MCG May 1 '19 at 16:01
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The calculation for 1k and 3k resistors looks wrong:

\$R_{eq}=\frac{1}{\frac{1}{1000}+\frac{1}{3000}}=\frac{3000*1000}{1000+3000}=750\Omega\$

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  • \$\begingroup\$ R8 is missing as well.. \$\endgroup\$ – Eugene Sh. May 1 '19 at 15:20
  • \$\begingroup\$ R8 got rolled into the 32k, the total for all three would be 30.75k \$\endgroup\$ – Voltage Spike May 1 '19 at 15:21
  • \$\begingroup\$ Well, then it can't have BC voltage \$\endgroup\$ – Eugene Sh. May 1 '19 at 15:22
  • \$\begingroup\$ I see your point \$\endgroup\$ – Voltage Spike May 1 '19 at 15:22
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First off, you incorrectly reduced the combination of 1k||3k to 2.35k. This is incorrect on it's face since whenever you have resistors in parallel, the combined resistance is less than either resistance. You should use the formula (1/R1 + 1/R2 + ... + 1/Rn)^-1 to reduce the parallel circuit. When this is applied you get a more reasonable value of 750.

Second, you mislabeled node C. On the top circuit it is the node between R6||R7 and R8. On the bottom circuit, it is tied to ground.

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