0
\$\begingroup\$

I have a parallel circuit with 50 parallel wires each connected to main cables (+ and -) on top and bottom side of the circuit. I want to calculate the total resistance which I know is R = R1/ total number of parallel wires.

Should I also include main cables in my calculation? If so, should I measure the length of the cable from the first "resistor"? Should I only calculate the main + cable or both?

Thank you!

\$\endgroup\$
  • 1
    \$\begingroup\$ Should you? Depends what you are going to do with information.. \$\endgroup\$ – Eugene Sh. Jul 19 '17 at 21:56
  • \$\begingroup\$ parallel wires twisted are about 120 Ohm in AC RF, yet not rated for grid transients Cable Z (f)is only used when prop delay is near or greater than wave period of interest The grid can be assumed to be ~0 Ohms at line f, so your model is incorrect. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 19 '17 at 22:21
  • 1
    \$\begingroup\$ @TonyStewart.EEsince'75: OP mentioned "main cables", not "mains cables". I suspect that the question is so basic that your comment is a tad over complex. \$\endgroup\$ – Transistor Jul 19 '17 at 22:29
  • \$\begingroup\$ The load impedance will be the sum of the cable capacitance and the inverse sum of all the 1/R loads divided into the total source+ sum R's. to get Req/(Req+Rs) as the impedance ratio. Cable capacitance is 30~100pF/m depending on gap or twists/m to max with coax \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 20 '17 at 0:29
  • \$\begingroup\$ Main cable is undefined in electrical terms \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 20 '17 at 0:32
1
\$\begingroup\$

I guess it depends on some information that we don't have.

The "main cables" (which hopefully for your safety aren't mains cables), may be significantly thicker than the other smaller cables and have a negligible resistance.

If you want to model the resistance exactly, go ahead and measure the main cables as well, and add it in series with the parallel mess of wires. Unless you are doing something unsafe, the resistance of these wires really shouldn't make a difference.

On the topic of safety: If you are doing something where you are drawing enough current that your putting 50 wires in parallel to solve the problem, that isn't the solution your looking for. While it will lead to a lower overall resistance, there will still be fairly high current going through each of the wires, and you don't want them to burn up. Any power dissipated by your makeshift resistors will turn into heat, and potentially a lot of it.

Please make safe decisions, especially if you aren't that experienced with electronics yet, and avoid high voltage until you know what your doing.

And if you insist on still going through with this stuff: please keep one hand behind your back at all times. High voltage can really kill you if you aren't careful.

\$\endgroup\$
  • \$\begingroup\$ One should worry about the current rather than the voltage ;) 1 amp and 1 volt circuit can be very deadly. \$\endgroup\$ – KingDuken Jul 19 '17 at 22:19
  • \$\begingroup\$ Yes, but most low voltage sources are pretty severely current limited. Also, higher voltages allow higher currents to go more places (like through your heart). It is reasonable to be concerned about working with high voltage systems. \$\endgroup\$ – Essaim Jul 19 '17 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.