# Will a loose connection cause an increase in amps?



## carryyourbooks (Jan 13, 2010)

got a trouble call today. lost a leg at the main fusible disconnect (200a, single phase). replaced fuse. power is on. 5600 square foot home, all elctric, no gas.

i could smell electrical burn at the exterior disconnect where the fuse was changed. i got out the old temp gun and found the top of each fuse at around 150 degrees each. current outside temp was 62 degrees.

load tested each leg. 170 amps in one leg and 100 on other. they had the meter and disco mounted outside on a pole. went underground to panel inside. 

located hot spot in sub panel. (2) breakers for furnace were at 110 degrees. removed breakers and found burned buss bar. breaker would not stay clamped on buss for obvious reasons. (breakers were qo, btw).

after i turned breaker off that were connected buss, i took another load test and checked the temp at the top of the outside fuses. heat had come way down. amps were around 80 amps on each leg.

will these loose breaker connections cause an increase in amp draw when they are energized?


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## donselec (May 7, 2011)

you proved it...:thumbsup:


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## carryyourbooks (Jan 13, 2010)

donselec said:


> you proved it...:thumbsup:


seemed obvious to me due to the increased resistance. just wanted to check my logic.


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## Big John (May 23, 2010)

I'm having trouble understanding: You shut off a couple breakers and saw a drop of 90A on L1 and 20A on L2. Aside from being very imbalanced, I don't get what strikes you as strange about that?

And I guess as a short answer: No a loose connection shouldn't cause an increase in current draw. It's high resistance, so if anything, the current would go down by a very small amount.


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## triden (Jun 13, 2012)

What Big John said. A high resistance creates heat because there is a large voltage drop across it. The current would have a tendency to decrease through a bad connection.


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## carryyourbooks (Jan 13, 2010)

i will know more tomorrow. we are replacing the panel with the burnt buss. we will retest after. i told the HO she needs to bump up to 400a if her 5600 sq ft home is all electric and no gas. the HO is about to add alot more landscape lighting. they have a pool. i know they have 3 ac's and most likely the same amount of furnaces. 

we'll see.


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## uconduit (Jun 6, 2012)

Don't engineers calculate heat dissipation in Watts=Current (amps) x Current (amps) x Impedance (ohms) ( W=I^2Z )?

The need for a solid connection is profound when you consider that if the resistance of the connections equals .0001 ohms it will dissipate 2.89 watts at 170 amps. If the connection is loose and is let's say .01 ohms between the breaker and the bussbar 289 watts would be dissipated, right? Toasty


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## 220/221 (Sep 25, 2007)

Big John said:


> : No a loose connection shouldn't cause an increase in current draw. It's high resistance, so if anything, the current would go down by a very small amount.



Now I am confused. 

Are you talking about current on the line side of the bad connection or the load side?

My logic says that more resistance = more amperage. Two toasters = 2X resistance = 2x amperage.

As usual, I'm probably just confused.


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## ElectricBrent (Jan 1, 2013)

220/221 said:


> My logic says that more resistance = more amperage. Two toasters = 2X resistance = 2x amperage.


Bad analogy, this case would be more like adding resistance in series with the load due to the loose connection. Current is an inverse function to resistance and total resistance increases when added in series.

I=E/R

Voltage stays the same so if resistance increases amperage decreases


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## Big John (May 23, 2010)

220/221 said:


> ...My logic says that more resistance = more amperage. Two toasters = 2X resistance = 2x amperage....


 That would definitely be true for parallel resistance, which is what you would get when you plug in two toasters: The more resistances you put in parallel, the lower your total circuit resistance will be.

But when it's in series, like a bad connection, the more resistance you put in, the higher your overall circuit resistance.


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## carryyourbooks (Jan 13, 2010)

as for now, i left breakers off where buss was burned when i left yesterday. load tested this morning. L1 had 20 amps at main and L2 had 16 amps. i had 6 breakers off. some were for ac's (energized, but not on yesterday). furnace was running too. temp at the main disconnect and panel were verry normal and well within range.


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## carryyourbooks (Jan 13, 2010)

interior panel swapped. rechecked load. most we could get with furnace running was 140 amps and balanced. 

the top side of the fuse was already at 180 degrees shortly after turning on the power. it was cool where the blade went into the clamp, but both fuses were hot only on top. i tested the lugs, wires, neutral, both hots, and the entire length of the fuse.

only hot on the top metal portion of the fuse? besides being undersized for the service, is there any other reason why it would get so hot in this one particular place and no where else?


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## MDShunk (Jan 7, 2007)

To add some confusion, in parallel conductors, a loose connection in one will raise the current measured in all the GOOD connections. The parallel conductor with the lowest amp draw is the one with the worst connection(s).


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## hardworkingstiff (Jan 22, 2007)

MDShunk said:


> To add some confusion, in parallel conductors, a loose connection in one will raise the current measured in all the GOOD connections. The parallel conductor with the lowest amp draw is the one with the worst connection(s).


Glad to see you post, I think some of us were wondering if you were OK. Now we know you are. :thumbsup:


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## Semi-Ret Electrician (Nov 10, 2011)

IMO, if a motor is being fed thru a loose connection the motor will slow (pull high amps) or stop (locked rotor) pulling really high amps. 

The loose connection will go into thermal runaway getting progressively worse, like a welders arc.


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## rrolleston (Mar 6, 2012)

IMHO any loose connection that causes heat is almost acting like a heating element that uses energy. It takes energy to create heat.


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## ibuildpower (Jan 8, 2013)

carryyourbooks said:


> got a trouble call today. lost a leg at the main fusible disconnect (200a, single phase). replaced fuse. power is on. 5600 square foot home, all elctric, no gas.
> 
> i could smell electrical burn at the exterior disconnect where the fuse was changed. i got out the old temp gun and found the top of each fuse at around 150 degrees each. current outside temp was 62 degrees.
> 
> ...


A loose connection has more resistance than a tight one. That resistance drops voltage and resists current flow. The voltage drop is disipated as heat and the heat is proportional to the current flow in the rest of the circuit up to the point when thermal runaway occurs. 

I think your circuit current would be less with the poor connection at the breaker that you describe, and I think you can prove it with ohm's law. Treat the poor connection as a series resistor and Thevenize the circuit (reduce the net resistance to a single combined resistance by adding them together). Calculate the power and current in the circuit both ways -- with and without the series resistance of the porr connection and I'll bet you find I'm right. 

Power and current should go down as resistance to current flow goes up.


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## uconduit (Jun 6, 2012)

Think of a simple DC circuit. Every conductor is a resistor, whether intentional or not. 

From the NEGATIVE power source to the load think of it as a SERIES of resistors. connector, cable, termination, switch, then finally to the LOAD.

From LOAD, it goes through a SERIES of resistance back to the POSITIVE.

R(LOAD+CONDUCTORS)=R(CONDUCTOR-)+R(LOAD)+R(CONDUCTOR+)

1.018Ω = .009Ω + 1Ω + .009Ω (adding series resistance)

just to mix it up a little lets just say that the humidity in the AIR is high enough for there to be 10MΩ of resistance in the circuit also "short circuited" between the positive and negative power source terminals 

1/R(total) = 1/R(load+conductors) + 1/R(air) (adding parallel resistance)
1/(1.0179998964Ω)=1/10MΩ + 1/1.018Ω


the AIR circuit, assuming 170VDC would only allow 17μA to flow.

the CONDUCTOR+LOAD+CONDUCTOR circuit would allow 166.994 amps to flow. Exactly 166.994 flows through the entire circuit, just like how the same current flows through every part of a 4-20mA instrumentation circuit.

So as long as the load is purely resistive the the current flow is at its highest when the circuit has conductor, termination resistance approaching zero and the current will always be less as the total resistance changes. However motors are not "purely" resistive and will get angry at undervoltage and draw more current so I guess it just depends. Also with AC circuits the term is impedance which IIRC is the vector sum of reactance and resistance.


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