There's some great answers here already, but I thought maybe showing some maths would help with understanding.
- To keep things a bit simpler, we're going to focus on purely resistive loads, and ignore impedance, power factor, etc.
- To get more accurate result, we'll include the resistance of all the wiring in the circuit. For all the examples, we'll assume 200' of 12 AWG copper wire is used (0.00193 ohms/ft). Unless otherwise specified
we'll start with a simple circuit, that contains only a single 60 watt light bulb
We can calculate the total resistance as such
Rt = R1 + R2
I this case,
R1 is the resistance of the light bulb, while
R2 is the resistance of the wire.
Rt = 240 ohms + 0.386 ohms
Rt = 240.386 ohms
Next, we can use Ohm's law to calculate the total current through the circuit.
It = E / Rt
It = 120 volts / 240.386 ohms
It = 0.499 amperes
Through this example, you can see that the circuit will only draw about half an ampere.
What would happen if we removed the light bulb, and "completed the circuit"?
With the light bulb gone, the only resistance in the circuit is the wire.
Rt = 0.386 ohms
Using that to calculate the current
It = 120 volts / 0.386 ohms
It = 310.88 amperes
We end up with a current draw 15.5 times the rated current (20 amperes) allowed by the circuit breaker. This causes the circuit breaker to trip, and open the circuit.
Next we'll take a look at a more complex example, where we have three bulbs in parallel.
To calculate the resistance in a parallel circuit, it's not as simple as adding the resistances together. Instead you have add the reciprocals, and divide 1 by the result.
Rt = 1 / (1/R1 + 1/R2 + 1/R3)
Rt = 1 / (1/240 ohms + 1/240 ohms + 1/240 ohms)
Rt = 80 ohms
Next we'll have to add in the resistance of the wire in the circuit.
Rt = 80 ohms + 0.386 ohms
Rt = 80.386 ohms
We can calculate the current through the circuit as.
It = 120 volt / 80.386 ohms
It = 1.49 amperes
Finally, let's remove one of the bulbs and "complete the circuit".
In this example, the bulb resistance will be replaced by the resistance of 1' of 12 AWG copper wire.
Rt = 1 / (1/240 ohms + 1/240 ohms + 1/0.00193 ohms
Rt = 0.001929969 ohms
Again we have to add in the resistance of the wire in the circuit.
Rt = 0.001929969 ohms + 0.386 ohms
Rt = 0.387929969 ohms
Due to the low resistance, we can assume the current will be quite high.
It = 120 volts / 0.387929969 ohms
It = 309.3341829759 amperes
Once again the current is more than 15.5 times the rated circuit capacity, which should hopefully trip the circuit breaker.
By "Completing the circuit", you're actually creating a short-circuit (low resistance path). Because the resistance through this path is so low, the current will always be quite high. The circuit breaker reacts to the high current, and opens the circuit before the wiring can be damaged.