Resistance and voltage drop

[Brendon Smith]

Hey guys, I’m still learning, and trying figure out the basics. I would love some help understanding this.. If Amps X Resistance = Volts, then why does increasing the resistance by using a longer or smaller conductor ACTUALLY lower voltage? Why is it actually the opposite of OHM’S Law in this scenario?


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[LarryFine]

Hey guys, I’m still learning, and trying figure out the basics. I would love some help understanding this.. If Amps X Resistance = Volts, then why does increasing the resistance by using a longer or smaller conductor ACTUALLY lower voltage? Why is it actually the opposite of OHM’S Law in this scenario?

It's the voltage available at the load end that reduces because of the circuit resistance.

Electrical/Electronic - Series Circuits

 

[LarryFine]

To add, let's say you have a (resistive) load rated at 120 volts and 10 amps. Ohm's Law tells us that the load's resistance must be (R = E/I = 120/10) 12 ohms. Now, let's say the circuit conductors adds 1.5 ohms on each line, or 3 ohms, for a total circuit resistance of 15 ohms.

So, with a total circuit resistance of 15 ohms, we can calculate that the resultant current is (I = E/R = 120/15) 8 amps, instead of 10 amps. Now, we can see that the voltage at the load is (E = IxR = 8x12) 96 volts, and the voltage drop is (E = IxR = 8x3) 24 volts, lost on the way.

While the load should theoretically be consuming (P = ExI = 120x10) 1200 watts, using "Watt's Law" tells us that (P = ExI = 96x8) 768 watts are being consumed by the load, while (P = ExI = 24x8) 192 watts are being wasted heating the conductors, a total of (768+192) 960 watts.

 

[Brendon Smith]

Thanks!


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[Brendon Smith]

To add, let's say you have a (resistive) load rated at 120 volts and 10 amps. Ohm's Law tells us that the load's resistance must be (R = E/I = 120/10) 12 ohms. Now, let's say the circuit conductors adds 1.5 ohms on each line, or 3 ohms, for a total circuit resistance of 15 ohms.

So, with a total circuit resistance of 15 ohms, we can calculate that the resultant current is (I = E/R = 120/15) 8 amps, instead of 10 amps. Now, we can see that the voltage at the load is (E = IxR = 8x12) 96 volts, and the voltage drop is (E = IxR = 8x3) 24 volts, lost on the way.

While the load should theoretically be consuming (P = ExI = 120x10) 1200 watts, using "Watt's Law" tells us that (P = ExI = 96x8) 768 watts are being consumed by the load, while (P = ExI = 24x8) 192 watts are being wasted heating the conductors, a total of (768+192) 960 watts.

That makes perfect sense, thank you for the great description!


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[ggunn]

That makes perfect sense, thank you for the great description!


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Think of it as a voltage divider. The total voltage drop across the series resistances of + conductor, load, and - connector has to equal the supply voltage; if you increase the voltage drop in the conductors by making them longer and/or smaller, there is less voltage left for the load.

 

[GoldDigger]

To summarize the critical point: Ohm's law tells you the relationship between the voltage across a resistance (between the two terminals of that resistance) and the current through that resistance. Not the voltage somewhere else in the circuit.

 

[LarryFine]

You have to start with knowing which values are the constants and which are the variables. For a normal circuit calculation, the supply voltage and the circuit resistance are known, and you calculate the resulting current.

Keep in mind things such as the difference between varying the voltage applied to a fixed load, and intentionally manufacturing the resistance of a load to be used on a different supply voltage for the same resultant power.

 

[ggunn]

You have to start with knowing which values are the constants and which are the variables. For a normal circuit calculation, the supply voltage and the circuit resistance are known, and you calculate the resulting current.

Keep in mind things such as the difference between varying the voltage applied to a fixed load, and intentionally manufacturing the resistance of a load to be used on a different supply voltage for the same resultant power.

And then you have PV systems, which are current sources instead of voltage sources like the grid and batteries. Voltage drop in the conductors turns into voltage rise at the terminals of a PV inverter.