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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
    Commented May 1, 2019 at 15:21
  • \$\begingroup\$ By simple observation, you should know that R11 will be something less than 31k. Thus, your simplification of R6 || R7 is faulty. \$\endgroup\$
    – qrk
    Commented Aug 26, 2021 at 4:15

5 Answers 5

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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\$ Commented May 1, 2019 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
    Commented May 1, 2019 at 15:50
  • \$\begingroup\$ Yes it is with respect to GND. I will try again. \$\endgroup\$ Commented May 1, 2019 at 15:55
  • \$\begingroup\$ When I get the answer from the divider equation, I have to subtract it from the source voltage right? \$\endgroup\$ Commented May 1, 2019 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
    Commented May 1, 2019 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.
    Commented May 1, 2019 at 15:20
  • \$\begingroup\$ R8 got rolled into the 32k, the total for all three would be 30.75k \$\endgroup\$
    – Voltage Spike
    Commented May 1, 2019 at 15:21
  • \$\begingroup\$ Well, then it can't have BC voltage \$\endgroup\$
    – Eugene Sh.
    Commented May 1, 2019 at 15:22
  • \$\begingroup\$ I see your point \$\endgroup\$
    – Voltage Spike
    Commented May 1, 2019 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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To apply the voltage divider rule, first you need a voltage divider, which is a simple chain of resistors connected in series, between a known potential difference.

The only things in your circuit which don't fit that description are the parallelled combination of R6 and R7, so the first thing to do is work out what single resistance can replace them. Once we do that, we'll have a chain of three single resistors, connected across a known potential difference of 6V, which is emminently suitable for application of the voltage divider rule.

Let's call \$R_6\$'s and \$R_7\$'s combined resistance \$R_X\$:

$$ R_X = \frac{1}{\frac{1}{R_6} + \frac{1}{R_7}} = \frac{R_6 \times R_7}{R_6 + R_7} = \frac{3k\Omega \times 1k\Omega}{3k\Omega + 1k\Omega} = \frac{3}{4}k\Omega = 750\Omega $$

Redrawing the circuit with \$R_X\$ in place, replacing \$R_6\$ and \$R_7\$ (I've also rearranged it a little, to help with visualising the coming arithmetic, but the circuit is identical in all other respects):

schematic

simulate this circuit – Schematic created using CircuitLab

\$ V_{R5} \$, \$ V_{RX} \$ and \$ V_{R8} \$ represent the voltages that will be present across resistors \$ R_5 \$, \$ R_X \$ and \$ R_8 \$ respectively. I will call the "absolute" voltage at node A \$ V_A \$, so at node B we have \$ V_B \$, and so on.

Let's apply the voltage divider rule, which says that the resistors will each develop a "share" of the total voltage across the entire chain, in proportion to their individual resistances. In other words, if we call \$ R_{TOTAL} \$ the combined restistance of the entire chain

$$ R_{TOTAL} = R_5 + R_X + R_8 = 5k\Omega + 750\Omega + 30k\Omega = 35.75k\Omega $$

then:

$$ V_{R5} = V1 \times \frac{R_5}{R_{TOTAL}} = 6V \times \frac{5k\Omega}{35.75k\Omega} = 0.8392V $$

$$ V_{RX} = V1 \times \frac{R_X}{R_{TOTAL}} = 6V \times \frac{750\Omega}{35.75k\Omega} = 0.1259V $$

$$ V_{R8} = V1 \times \frac{R_8}{R_{TOTAL}} = 6V \times \frac{30k\Omega}{35.75k\Omega} = 5.035V $$

Just for completeness, let's also apply Kirchhoff's Voltage Law (KVL), to work our way up the chain, and identify the "absolute" voltages at each node on the way. Starting at D, the voltage there is 0V, as indicated by the ground symbol:

$$ V_D = 0V $$

We know that the potential at point C must be higher than at D by the amount across \$ R_8 \$

$$ V_C = V_D + V_{R8} = 0V + 5.035V = 5.035V $$

By that same logic we continue up the chain to find \$ V_B \$ and \$ V_A \$. The potential at point B must be higher than at C by the amount across \$ R_X \$:

$$ V_B = V_C + V_{RX} = 5.035V + 0.1259V = 5.161V $$

$$ V_A = V_B + V_{R5} = 5.161V + 0.8392V = 6.000V $$

This last value for \$ V_A \$ must be 6V, because we've arrived at the node connected to the top of the battery, which is obviously 6V higher than the node at the bottom of the battery, at 0V. KVL for the win.

Viewing a voltage divider like this in the context of KVL, clearly the voltages across the individual resistances \$ V_{R5} \$, \$ V_{RX} \$ and \$ V_{R8} \$ must add up to equal the voltage source across them all, if KVL is to be obeyed. You can use this as a simple check of the validity of your arithmetic.

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Regarding to your calculation on parallel resistors:

If one does not have calculator or notebook along while figuring out parallel resistors total resistance, and values are somewhat easy to multiply, there's easy trick for them:

If you have 1k and 3k parallel

  1. Take the biggest value 3k
  2. Count how many times each resistor value fits into that 3k. So 1k fits 3 times, and 3k fits 1 time.
  3. Now divide that step 1 value, 3k with step 2 sum, 4 We get 3k/4 which is 750R

Another example with 1k, 2k, 5k parallel:

  1. take biggest, 5k
  2. count how many times each value will fit into that 5k. 1k x5, 2k x2.5, 5k x1
  3. result is 5k/8.5 which is 588R

It's the same math but put into easier way if you lack fancy calculator.

And remember that when calculating parallel resistors total value, the result is always smaller than the smallest resistor in the group (pointing at the erroneus simplified circuit).

Bonus example 1k, 3k, 4k, 10k, 50k parallel:

  • 50k / (50+16.7+12.5+5+1) = 50k / 85.2 = 587R

Note that the above isn't the right way, but it's a good way if you don't have calculator with you.

And when you finally get to the voltage calculations:

  1. Solve parallel resistors first
  2. Rtotal = R1 + R2 + R3 ...
  3. Itotal = V/Rtotal
  4. Voltage drops over resistors like this: VR5 = R5 * Itotal
  5. And if you need the current over invividual parallel resistors, you already have voltage drop over them and their single resistor values. IRx = VRx / Rx
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