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I'm trying to get a better understanding of the voltage and current dividers.

The question for this circuit is:

What are the voltages and current through all the resistors?

I calculated the total current that flows through the circuit, but now I'm stuck with this question. I'm not asking for the answers but I need a plan to tackle this problem. In the attachment you can find the circuit with all the information.

Circuit

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  • \$\begingroup\$ You basically have to solve simultaneous equations. You can use Ohm's Law, KVL, and KCL to come up with equations. You should be able to find enough equations that you can solve them to get a unique solution, and they should not be too difficult to solve (especially if you also use the series/parallel resistor laws) \$\endgroup\$
    – user253751
    Sep 14, 2020 at 17:19
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    \$\begingroup\$ Hint: What is the current through R1? Then, what is the voltage across R1? Then, what is the voltage across R2? Then, what is the current through R2? \$\endgroup\$
    – user253751
    Sep 14, 2020 at 17:19
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    \$\begingroup\$ The current that flows trough R1 is the total current of the circuit? Or am I wrong? \$\endgroup\$ Sep 14, 2020 at 17:40
  • \$\begingroup\$ @ArieHoekstra the statement voltage through a resistor is incorrect, it is voltage across a resistor and current through a resistor \$\endgroup\$
    – jsotola
    Sep 14, 2020 at 17:41
  • \$\begingroup\$ @ArieHoekstra yes and you said you figured out what that current was. \$\endgroup\$
    – user253751
    Sep 14, 2020 at 17:59

4 Answers 4

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You can find equivalent resistance of each branch. Replace resistors with equivalent resistors until you have only one resistor left. Calculate voltage and current and then start unwinding again, as you break Req in its components you will find voltage and current at each resistor.

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Whenever I deal with parallels I think "conductivity" or "conductance" G instead of "resistance" R. G = 1/R. Unit is "siemens" 1 S or "mho" $$1 \text{S} = 1 Ω^{-1}$$.

  • In parallel paths the conductances are additive.
  • In sequential paths the resistances are additive.
  • In parallel paths all the voltages are equal.
  • In sequential paths all the current (amperages) are equal.

I hate to memorize complex formulas of Kirchhoff's laws, I do it all from these basic principles. If in doubt, I think water buckets and garden hoses.

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So you say you calculated the total current.

The voltage across R2 is U1 - I*R1. Subtract the current through R2 and you know the current through R3. Repeat until you get to R7 + R8 and you will have all the currents and voltages.

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Start from right: the old fashionend way without a calucator 2 + 2 = 4 4 // 4 = 2 4 + 2 = 6 4 // 6 = 24/10 = 12/5 8 + 12/5 = 52/5 4 // 52/5 = 26/9 2 + 26/9 = 44/9

19 5/9 = 176/9 / 44/9 = 4 A Ur1 = 2 Ohm * 4 A = 8V Ur2 = 19 5/9 - 8 = 11 5/9 V Ir2 = 11 5/9 / 8 = 13/9 A and so on...

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  • \$\begingroup\$ Thanks for your comment. But can you explain the current trough R2? I thought it was 11 5/9 / 4. \$\endgroup\$ Sep 15, 2020 at 7:04
  • \$\begingroup\$ Sorry Arie.. I made a failure!! Ir2 = 11 5/9 / 4 = 26/9 A \$\endgroup\$
    – Bill
    Sep 17, 2020 at 19:06

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