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Just one to clear up the nonsense

I’ve been told that when a loose connection happens it causes a high resistance joint causing voltage drop and heating of the joint. My question is will this cause a breaker to trip. I’ve been told before that due to the added resistance the circuit then draws more current which will trip the breaker but I know this isn’t true as the power is determined by voltage x current so the added resistance will limit the current so what would trip the breaker (connection being down stream at the load)

thanks guys again for your explanations
 
The amount of heating will be based on the current flowing in the circuit and the resistance of the bad joint itself. The joint will not in itself draw current, so if the load is not enough to trip the breaker a loose connection will not suddenly increase the current flowing.

The only situation I can think of where this may not hold true is if the load profile is such that there is a significant in-rush current. If the bad joint is such that the supply to the load is fluctuating, if the timing was right, it may be possible for the load to draw it's in-rush current much more frequently than expected which may have the potential to overload the OCPD. If the load is purely resistive, this cannot happen (think about resistors in series and Ohms law).
 
A bad connection adds to the resistance of the circuit and usually reduces the current, in which case nothing should trip (unless it is arcing in which case it could trip an AFDD). But in many cases the actual change in current is tiny. Let's look at these effects in more detail.

How does a bad connection influence a heater circuit?

Suppose that a loose connection to a 230V 3kW immersion heater starts to increase in resistance until it is dissipating the heat of a small soldering iron (15W). The normal current was 3000 / 230 = 13A and the element resistance (230)² / 300 = 17.6Ω. The resistance of the bad connection adds only approximately 15 / (13)² = 0.089Ω to the total circuit resistance, so the decrease in current is approximately 0.089 / 17.6 = 0.5%. I.e. there is no way you would know from taking measurements that anything was wrong.

The crux of this is that even though the power dissipated in the bad connection is a small fraction of the load power, the connection can reach a dangerously high temperature because the heat is focused in a tiny spot and unable to dissipate freely as it can in the heating element.

If we consider a situation where the current reduces by 50%, clearly the total resistance has doubled, so the resistance of the faulty connection must be equal to the element resistance of 17.6Ω. The total power dissipation is now (230)² / 35.2 = 1500W, of which half is dissipated in the element and half in the connection. With 750W (the heat of fifty small soldering irons) the connection will soon reach red heat and self destruct, so this is not a situation that will normally be encountered for long.

How does the current drawn by a load vary with varying voltage?

Your original premise was that reducing the voltage (due to the bad connection) increases the current drawn by a load. As we saw above, that is not true of a resistive load which obeys Ohm's law: decrease the voltage and the current decreases in proportion.

But not all loads are resistive. Consider a switched-mode power supply, like the one in the computer you're using and most electronic goods. This is fully regulated - it delivers a constant voltage to the guts of the computer and that takes whatever power it needs. The power supply then draws whatever current it needs to achieve that power. Because P=VI, if the voltage decreases the current increases in inverse proportion.

Other kinds of load react in yet different ways to decreasing voltage. Induction and synchronous motors typically draw more current, universal motors typically draw less. Filament lamps draw less but not as much less as the heating element, due to their positive temperature coefficient of resistance.

In summary we note that reducing the voltage can have a wide range of effects on the current according to the type of load, anywhere between proportionality (resistive loads) and inverse proportionality (constant power loads).

However there is an important point to take away when considering the bad connection scenario: No matter what kind of load is concerned, the current will not increase more than is needed to provide the extra power dissipated at the connection, which as we have seen is often a small fraction of the load power. Therefore, it is not normally possible for OCPD to detect the presence of the bad connection.

What about those other failure modes that could cause an overcurrent trip?

There were two possibilities mentioned above. One is the repeated inrush of a load where an intermittent connection is arcing and interrupting the supply to a load. This is unusual but it can and does happen. We had a thread IIRC earlier this year where a bad connection in a lighting circuit was causing the MCB to trip on the inrush pulses of some LED drivers.

The other case was a bad connection in one line of a 3-phase motor supply, drastically increasing the current in the other two lines. This is a slightly different scenario because it is not the current through the bad connection itself that increases - rather it is forcing the motor to obtain its input power via the remaining good connections. This is a valid and significant problem as it can cause rapid overheating and damage to the windings. Therefore motor overload relays can be designed to trip when the line currents are unbalanced, even before any of them get to the normal overcurrent trip threshold. A similar effect can occur with a 3-phase rectifier or SMPSU input.

TL, DR; No, a bad connection cannot normally increase the current in a circuit and therefore does not cause OCPD to trip, except in a couple of special cases.
 
Last edited:
We had a thread IIRC earlier this year where a bad connection in a lighting circuit was causing the MCB to trip on the inrush pulses of some LED drivers.
 

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