Current Divider

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PhaseShift

Senior Member
I have the following circuit with R1 and R2 in parallel

100V
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R1 R2
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Gnd Gnd

If R2 is a short circuit then all current will flow through R2 and no current through R1 therefore no voltage across R1.

Even a short circuit has some small impedence, so if R2 is a short circuit with .00001 ohm impedence then based on the current divider wont some current still flow through R1 and thus casue a 100V voltage drop across R1? Wont current divide i porportion and cause enough current through R1 to cause 100V drop?
 
In the example circuit that you give, the current through R1 is equal to 100V/R1, and the value of R2 has no effect on the current through R1. If you put a series resistor between the 100V supply and the parallel R1/R2 branches, it is a completely different question with a completely different answer.
 
SAC said:
In the example circuit that you give, the current through R1 is equal to 100V/R1, and the value of R2 has no effect on the current through R1. If you put a series resistor between the 100V supply and the parallel R1/R2 branches, it is a completely different question with a completely different answer.
Click to expand...

Would the R1 and R2 parallel resistors not be a current divider with the ratio of resistances determining the current through each resistor?
 
PhaseShift said:
Would the R1 and R2 parallel resistors not be a current divider with the ratio of resistances determining the current through each resistor?
Click to expand...
Yes.

But they don't divide the original pre-fault current.

Suppose R1=R2=10 ohms. Then you have 20 amps supplied, 10 through each resistor.

No Let R2 drop to 0.00001 ohms. The current through R2 is now 10,000,000 amps. This does not change the current through R1 as it is still 10 amps. The total current is now 10,000,010 amps.
 
SAC said:
In the example circuit that you give, the current through R1 is equal to 100V/R1, and the value of R2 has no effect on the current through R1. If you put a series resistor between the 100V supply and the parallel R1/R2 branches, it is a completely different question with a completely different answer.
Click to expand...
And more akin to a real life situation.
 
mivey said:
No Let R2 drop to 0.00001 ohms. The current through R2 is now 10,000,000 amps. This does not change the current through R1 as it is still 10 amps. The total current is now 10,000,010 amps.
Click to expand...

But wont the 10A still going through R1 cause a voltage drop of 100V across R1? I thought that when a resistor or group of parallel resistors was shorted out, that the voltage drop across these resistor(s) was 0V?
 
PhaseShift said:
But wont the 10A still going through R1 cause a voltage drop of 100V across R1? I thought that when a resistor or group of parallel resistors was shorted out, that the voltage drop across these resistor(s) was 0V?
Click to expand...
You can't have it both ways. Given the resistors you have shown, you can't have one voltage across R1 and a different one across R2.

If the voltage drop across R1 goes to zero, the voltage drop across R2 goes to zero. In the case of the short, if you have a 100 volt drop across R1, you will have a 100 volt drop across R2.

With the circuit shown, if you short R2 and make it a near zero ohm resistance, it will still have a 100 volt drop across it because the supply will pump out near infinite current.

This is not what happens in the real world, of course, as the supply will pump out a high initial fault current, but be dragged down (except for a flux capacitor). Also in the real world, there are other impedances in the circuit that have their own voltage drops.

add: JFTR, my post #4 should have read: "Now Let R2 drop to 0.00001 ohms"
 
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mivey said:
You can't have it both ways. Given the resistors you have shown, you can't have one voltage across R1 and a different one across R2.

If the voltage drop across R1 goes to zero, the voltage drop across R2 goes to zero. In the case of the short, if you have a 100 volt drop across R1, you will have a 100 volt drop across R2.

With the circuit shown, if you short R2 and make it a near zero ohm resistance, it will still have a 100 volt drop across it because the supply will pump out near infinite current.

This is not what happens in the real world, of course, as the supply will pump out a high initial fault current, but be dragged down (except for a flux capacitor). Also in the real world, there are other impedances in the circuit that have their own voltage drops.

add: JFTR, my post #4 should have read: "Now Let R2 drop to 0.00001 ohms"
Click to expand...


So you are saying that If I had 5 resistors in parallel between a 100V source and ground and one of these resistors shorted out that I would still read 100V across all of the resistors?

I thought I had saw somewhere that when a resisotr was shorted out the voltage across all resistors went to zero?

If one of the 5 resistors went to ground than would'nt both sides of the remaining resistors be at the same ground potential?
 
PhaseShift said:
So you are saying that If I had 5 resistors in parallel between a 100V source and ground and one of these resistors shorted out that I would still read 100V across all of the resistors?

I thought I had saw somewhere that when a resisotr was shorted out the voltage across all resistors went to zero?

If one of the 5 resistors went to ground than would'nt both sides of the remaining resistors be at the same ground potential?
Click to expand...
You can put in as many resistors as you want. As long as the power supply could supply enough current, it will try to maintain 100 volts at the output terminals (across the resistors).

You have to cross the line between theory and reality at some point. In reality, when you short the resistor, you have shorted out the supply and the voltage across the other resistors will drop to zero because the power supply is at zero (this is due to things internal to the source).

You won't have zero volts across one resistor, and a 100 volt drop across the others.

We are talking about parallel resistors. You may be confusing this with how a circuit with series resistors would react. If R1 and R2 were the same size, and in series across the power supply, they each would have a 50 volt drop across them. If you shorted R2, then you would have zero volts across R2, and 100 volts across R1.

add: as long as you could supply enough current to keep the voltage at the supply terminals at 100 volts.
 
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drbond24 said:
Yes.

In order for the voltage across the resistor to go to zero, you would need an open, not a short.
Click to expand...

Opps, that's backward. All the voltage applied to a circuit appears arcoss an open circuit, and 0 volts appears across a short circuit.

Steve
 
steve66 said:
Opps, that's backward. All the voltage applied to a circuit appears arcoss an open circuit, and 0 volts appears across a short circuit.

Steve
Click to expand...
I am thinking he meant an open between the supply and the resistor(s).
 
PhaseShift said:
I thought I had saw somewhere that when a resisotr was shorted out the voltage across all resistors went to zero?
Click to expand...

That's generally true.

For the special case where the parallel set of resistors is the entire circuit load, then you need to consider a real power supply. It would have a small internal resistance.

Imagine your same setup, with the short across your load resistor, and something like a 0.1 ohm internal resistance in the battery.

Then the voltage across the parallel resistors will go to zero when you short them out, and the entire battery voltage will be dropped across the batterys internal resistance.

If you then imagine your short across the parallel resistors to be 0.00001 ohms, you will find that you still have a very small (almost zero) voltage across the paralllel resistors.

Steve
 
mivey said:
You have to cross the line between theory and reality at some point. In reality, when you short the resistor, you have shorted out the supply and the voltage across the other resistors will drop to zero because the power supply is at zero (this is due to things internal to the source).
Click to expand...

Maybe this part I am missing. So because of source impedence it will cause voltage across parallel resistor network to be 0V?

What if there was a large resistor between the 100V supply and the top terminals of the two resisotrs?

Say if stray voltage goes to a relay coil. You shunt out relay by shorting across it to ground to bleed of voltage. Doesn't voltage still drop across coil?
 
PhaseShift said:
Maybe this part I am missing. So because of source impedence it will cause voltage across parallel resistor network to be 0V? Yes. As long as the source has not been dragged down, the supply voltage will drop across the source impedance. In reality, the short will have some impedance and the source (because of "stiffness" will probably pump enough current into it to burn something up.

What if there was a large resistor between the 100V supply and the top terminals of the two resisotrs? You would have to have a high supply voltage to overcome this resistance and still supply 100 volts to R1 and R2, or you would have a smaller voltage across R1 & R2. A more realistic source impedance is a small resistor. See below

Say if stray voltage goes to a relay coil. You shunt out relay by shorting across it to ground to bleed of voltage. Doesn't voltage still drop across coil? The voltage across the coil will be the same as the voltage across then shunt
Click to expand...
Let's have a large resistor of 100 (R0) to go with the 10 ohms for R1 & R2. Now you have:

.................I.........I2
..__R0___________>_________>__
.|........+.............|.....|
Vs.......100V...........R1....R2
.|........-.........I1.\|/....|
.|______________________|_____|

Initially, you have:
I1=I2=100/10=10 amps
I=I1+I2=20 amps
Vs=100+R0*I= 2100 volts

Now short R2:
V_R1=V_R2=0
I=Vs/R0=21 amps
I1=I*R2/(R1+R2)=0
I2=I*R1/(R1+R2)=21 amps




Now let Vs=100 volts:
.................I.........I2
..__________R0___>_________>__
.|..+...................|.....|
Vs.100V.................R1....R2
.|..-...............I1.\|/....|
.|______________________|_____|

Initially, you have:
I=Vs/(R0+R1||R2)=100/105=0.952 amps
V_R0=Vs-I*R0=100-100*0.952=95.24 volts
V_R1=V_R2=V=V_R0= 4.76 volts
I1=I2=V _R1/R1=0.476 amps

Now short R2:
V_R1=V_R2=0
I=Vs/R0=1 amp
I1=I*R2/(R1+R2)=0
I2=I*R1/(R1+R2)=1 amp




Now let's try a little more realistic one where R0 is small, say 0.01 ohms:
.................I.........I2
..__R0___________>_________>__
.|........+.............|.....|
Vs.......100V...........R1....R2
.|........-.........I1.\|/....|
.|______________________|_____|

Initially, you have:
I1=I2=100/10=10 amps
I=I1+I2=20 amps
Vs=100+R0*I= 100.2 volts

Now short R2:
V_R1=V_R2=0
I=Vs/R0=10,020 amps
I1=I*R2/(R1+R2)=0
I2=I*R1/(R1+R2)=10,020 amps

Add: It might be worth noting: Essentially, the smaller R0 becomes, the more "stiff" the system becomes and the more fault current it can supply.
 
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steve66 said:
That's generally true.

For the special case where the parallel set of resistors is the entire circuit load, then you need to consider a real power supply. It would have a small internal resistance.

Imagine your same setup, with the short across your load resistor, and something like a 0.1 ohm internal resistance in the battery.

Then the voltage across the parallel resistors will go to zero when you short them out, and the entire battery voltage will be dropped across the batterys internal resistance.

If you then imagine your short across the parallel resistors to be 0.00001 ohms, you will find that you still have a very small (almost zero) voltage across the paralllel resistors.

Steve
Click to expand...


I think I understand now.

If we ingonre source impedence for the sake of conversation then even with a parallel resistor shorted all 100V will still drop across parallel resistors, and all parallel voltages will be the same.

In reality, the source will have some small impedance. If we short one of the parallel resistors, then all the voltage will be dropped across source impedence and almost very little or zero voltage will be dropped across parallel resistors although all parallel resistors will have same voltage across them.
 
PhaseShift said:
I think I understand now.

If we ingonre source impedence for the sake of conversation then even with a parallel resistor shorted all 100V will still drop across parallel resistors, and all parallel voltages will be the same.

In reality, the source will have some small impedance. If we short one of the parallel resistors, then all the voltage will be dropped across source impedence and almost very little or zero voltage will be dropped across parallel resistors although all parallel resistors will have same voltage across them.
Click to expand...
You've got it.
 
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