Why does the amperage go real high if there is a low voltage short?

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Why does the amperage go real high if there is a low voltage short?
I used to be an HVAC technician. One time I had a service call in which there was no power to the thermostat. The thermostat did not have power because the fuse in the air handler was blown. The fuse in the air handler was blown because there was a low voltage short. The rubber coating on one of the thermostat wires was chewed off by a rodent. The exposed metal in the thermostat wire was touching the metal cabinet of the air handler. This was a low voltage short. This low voltage short caused the amp draw to go so high that the fuse blew. Why did the low voltage short of the exposed metal of the thermostat wire touching the metal cabinet of the air handler cause the amp draw to go so high?

I'm not an electrical engineer like you people. Please explain this to me like I am a five year old.
 
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That's completely normal.

Volts and amps are related by the expression V=IR, OR to reverse it, I = V/R, SO ... if the resistance is very very low, then the current has to be very very high (given that the power source HAS the power to deliver a lot of current).
 
phinds said:
That's completely normal.

Volts and amps are related by the expression V=IR, OR to reverse it, I = V/R, SO ... if the resistance is very very low, then the current has to be very very high (given that the power source HAS the power to deliver a lot of current).

What are you telling me? Are you trying to tell me that the reason that the low voltage short caused the fuse to blow is that the low voltage short caused the resistance to be a lot lower, making the amp draw a lot higher?
 
sevensages said:
... the low voltage short caused the resistance to be a lot lower, making the amp draw a lot higher?
Yes. Ohm's Law. I = V / R. Current is equal to voltage, divided by resistance.

The main AC supply distributes a standard voltage. The circuit presents a resistance, (or an impedance), that will regulate the current drawn, to deliver only the power needed from the AC grid. I = V / R ; Power, W = V * I.

A short circuit, is a fault condition, where a low-resistance path allows more current to flow than the wires can safely carry. A sacrificial (replaceable) fuse is used on each main AC circuit, to protect the wires from high-current fault conditions, by melting first.
 
sevensages said:
What are you telling me? Are you trying to tell me that the reason that the low voltage short caused the fuse to blow is that the low voltage short caused the resistance to be a lot lower, making the amp draw a lot higher?
No, I am not TRYING to tell you that, I am TELLING you that. Can you not follow the math?
 
phinds said:
No, I am not TRYING to tell you that, I am TELLING you that. Can you not follow the math?

I am not an engineer like you. In the OP, I asked you to explain it to me like I am a five year old.
 
A short, by definition, is a low resistance path for the electricity to take. Electricity prefers paths of low resistance.
 
sevensages said:
I am not an engineer like you. In the OP, I asked you to explain it to me like I am a five year old.
Well, 5 year old, think of it by way of the water analogy. If you have a narrow pipe, only a limited amount of water can flow through it. The pipe resists the water flow by constricting it like the resitance of an electrical circuit resists the flow of electrons. If you replace the narrow pipe with a HUGE pipe (equivalent to a short circuit for current flow) then the amount of water that can flow is basically only limited by the water supply (the voltage/power).
 
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