Is it OK to write the resistance in parallel like this?

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Writing the equation for resistance in parallel is acceptable, and the method of using the reciprocal of the sum of the reciprocals is a common mnemonic. Pocket calculators simplify this process significantly compared to slide rules. It is important to ensure that the total resistance in a parallel network is always less than the smallest resistor present. The same formula for calculating total resistance can also be applied to inductors, while capacitors require a different approach. Overall, the discussion emphasizes the practicality and elegance of the parallel resistance formula.
TBert
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I am creating an equations sheet for electronics forumlas and am wondering if it is OK to write the equation for resistance in parallel like this:
33bnUFo.png
 
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TBert said:
I am creating an equations sheet for electronics forumlas and am wondering if it is OK to write the equation for resistance in parallel like this: View attachment 197492
Yes.
 
My memory rhyme was always "Reciprocal of the sum of the reciprocals"
which when pocket calculators came along in 1970's became quite handy.

Enter R,, press 1/x, press M+, repeat until all R's accounted for, press Recall Memory, press 1/x .

That's one place where a pocket calculator really outdoes a slide rule .

Your formula is much more elegant. Just make sure there's not a R0 !

have fun,.

old jim
 
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A bit off track but in // r's the total network R should always be smaller than the smallest resistor in ohms.

Also, a network has no P/S whereas a circuit has a power supply.

And again what Jimbo said... R total = the sum of the reciprocals, reciprocated. You can use the same formula for inductors as well, whereas capacitors you treat opposite.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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