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I hope it's okay if I ask it here, but I've been playing with this current for quite some time now, and I've reached a dead end.

I am supposed to find the amperage and voltage of a specific resistor (Rx) only through simplification (so putting multiple resistors to one - no Thevenin or anything like that). I have done some work on the simplification, but at this point, I've no idea what to do next.

I am new to all this, so sorry if it's just somehow obvious...I just don't see it... I desperately tried to put it to circuits.js - I know the value I should get, but not how to get it. I know the formulas for merging two parallel and in-series resistors (+ofc the Ohm's law) but I can't see how to use it now without "merging" the Rx.

Here is how it looked like at the start, if anyone wants to see that. I am not really asking anyone to do the calculations and all, I just need to know what to calculate...

Thank you for going through this post, and big thanks in advance for any help.

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  • \$\begingroup\$ You'll have to simplify further, then un-simplify, I believe. \$\endgroup\$
    – Hearth
    Nov 6 '21 at 21:45
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The general strategy for problems like this is to combine resistors until you are left with a single resistor in parallel with the voltage source. Then you can use Ohm's Law to find the current through that single resistor.

Now that you know the current, take that value back to your next-to-last circuit simplification. Use the newly found current to find the voltage across the resistors in your next-to-last simplification.

Take the newly calculated voltages back one more step in the simplification process and find some resistor currents.

Keep doing this...calculating some new voltage or current at each step and transferring that back to an earlier step in the simplification...until you have the needed information.

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The left side with the power supply and Ra & R8 can be simplified by creating the thevenin equivalent circuit. Ra & R8 are really in series. And you can apply thevenin equivalent circuit again on Ra8 and Rb7, which are really on parallel. That should be able to give you just resistors in series.

If you have issues seeing this check that your Thevenin Equivalent circuit by measuring the open circuit voltage and the closed circuit current .

Or reduce it to this

enter image description here

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  • \$\begingroup\$ So how does that procedure compare to what the OP has actually done in their notes? \$\endgroup\$
    – TonyM
    Nov 6 '21 at 21:58
  • \$\begingroup\$ The OP said they can't use Thevenin for this problem. \$\endgroup\$ Nov 6 '21 at 22:00
  • \$\begingroup\$ I am trying to explain that for a resistance analysis we can "short"out the voltage supply to do his series and parallel resistor calculations \$\endgroup\$ Nov 6 '21 at 22:04

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