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I'm trying to prepare for FE exam. I'm having difficulty with this seemingly simple problem:

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  1. The solution calculates the node voltage and not voltage across 10k ohm resistor. Don't I have to find the individual current and then calculate the drop?
  2. The solution applies KCL separately for 4k and 10k. I can't logically refute this but when I put the two resistors in parallel, I don't get the same combined current going through both.
  3. My initial attempt at the problem was putting 4k and 3k in parallel. So (12/7) as the resistance. Then I used current divider:

I_10k = Isource (12/7) / [(12/7)||10k]

Then I calculated the voltage drop ,but I don't get anything even close to -14.6. I got like -0.1.

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2 Answers 2

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Hint: the three resistors are in parallel.

You can just work out the equivalent parallel resistance and then work out the voltage using Ohm's Law. There are only two nodes in the circuit - the top rail and the bottom rail. Every resistor sees the same voltage.

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  • \$\begingroup\$ I see so I did get voltage at node of 14.63 V. But I'm just confused between this concept of voltage across a resistor and a voltage drop. Let's say you have a voltage source of 4 v and two resistors in parallel that are 2 and 1. Wouldn't the voltage drop be different from voltage across the nodes? \$\endgroup\$
    – J L
    Commented Jan 23, 2022 at 13:30
  • \$\begingroup\$ No. Put one probe of your multimeter anywhere along the bottom rail and the other anywhere along the top and you'll get the same voltage. Therefore each resistor must have the same voltage across it. The current through each resistor will be inversely proportional to its resistance. \$\endgroup\$
    – Transistor
    Commented Jan 23, 2022 at 13:38
  • \$\begingroup\$ Ah I seeI think where I got the misunderstanding. Thank you \$\endgroup\$
    – J L
    Commented Jan 23, 2022 at 13:39
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  1. There is only 1 node so the voltage across the 10k resistor is the node voltage.
  2. When you put 2 resistors in parallel, you calculate the total current going into both. That current will divide between them according to the resistance of each one.
  3. You applied the current divider formula incorrectly. Recheck your calculation. Hint: the error is in the denominator.
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