Reply to the OP
Reply to the OP
Here are the numbers to explain what will happen in both Scenarios...
(to keep things simple, we will use a 1 Phase center tapped Transformer for these examples - such as used in most Residential areas)
BTW, symbols are:
- "E" = Voltage,
- "I" = Amperes,
- "R" = Resistance (across a fixed Resistance),
- "P" = True Power (Watts).
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* Scenario #1:
Two (2) 100 Watt, 120V Incandescent Lamps are used,
* Scenario #2:
One (1) 60 Watt 120V Incandescent Lamp + One (1) 200 Watt 120V Incandescent Lamp are used.
***SCENARIO #1***
Breaking the Common Neutral's connection between the Lamps' Circuit and the Transformer will result in both Lamps operating properly, without one - or both of the Lamps failing.
Description:
100 Watt, 120V Lamp (Incandescent) Data:
P = 100 Watts,
E = 120 VAC,
I = 0.8334 Amps (rounded up),
R = 143.9884 Ohms (rounded up).
If the Circuit is connected normal (no open neutral), the values across each Lamp will be:
E = 120V,
I= 0.8334 Amps.
The total Resistance per circuit (L-N) will be 143.9884 Ohms on "Each Side"
(Line 1 to N = 143.9884 Ohms, Line 2 to N = 143.9884 Ohms).
Each Lamp will have 0.8334 Amps pushed across the Filament by 120 VAC, with a resulting Power drawn of 100 Watts.
If the Common Neutral is opened between the Transformer and the Lamps' circuit (Open Neutral), the circuit values will now be:
Total Circuit E = 240 VAC,
Total Circuit R = 287.9768 Ohms
E ? R = 0.8334 Amps
Total Circuit I = 0.8334 Amps
Per-Lamp:
With 0.8334 Amps flowing through a fixed Resistance of 143.9884 Ohms, the resulting Voltage (potential difference) across the filament of each Lamp is 120 VAC:
R ? I = E
143.9884 ? 0.8334 = 120 Volts
As with the first example (normal circuit) each Lamp will have 0.8334 Amps pushed across the Filament by 120 VAC, with a resulting Power drawn of 100 Watts:
E ? I = P
120 ? 0.8334 = 100 Watts
100 Watts is the Lamp's rating, so the Lamp does not become damaged from excessive True Power draw.
***SCENARIO #2***
Lamp values:
* 60 Watt Incandescent Lamp:
E = 120 VAC,
P = 60 Watts,
I = 0.5 Amps,
R = 240 Ohms.
* 200 Watt Incandescent Lamp:
E = 120 VAC,
P = 200 Watts,
I = 1.667 Amps (rounded),
R = 76.85 Ohms (rounded)
In a "Normal" 3 Wire Circuit, each Lamp's filament has a Potential Difference of 120 VAC across it.
The 60 Watt Lamp allows 0.5 Amp to flow across it, and the 200 Watt Lamp allows 1.667 Amps to flow across it.
The common neutral will have 1.17 Amps flowing between the Lamp circuit and the Transformer.
Each Lamp will draw the rated True power (Watts) from the source, and therefore will operate properly without damage from excessive Power drawn through the filament.
If the Common Neutral of this 3 wire circuit is broken between the Transformer and the Lamps' circuit, the following will result:
* Total Circuit Voltage = 240 VAC,
* Total Circuit Resistance = 311.85 Ohms,
* Total Circuit Amperes = 0.7696 Amps,
* Total Circuit True Power = 184.7 Watts.
The 60 Watt Lamp will have the following values across its filament:
R = 240 Ohms,
E = 184.7 Volts,
I = 0.7696 Amps,
P = 142.15 Watts.
The 200 Watt Lamp will have the following values across its filament:
R = 71.85 Ohms,
E = 55.3 Volts,
I = 0.7696 Amps,
P = 42.56 Watts.
The 60 Watt Lamp will fail rapidly, due to the excessive True Power drawn from the supply, and "Dumped" into the Lamp's filament.
The True Power (Wattage) has exceeded the Lamp's rating by >300%.
The 200 Watt Lamp will be very dim, since the True Power drawn from the supply and 'Dumped" into the filament will only be 42.56 Watts.
The above Scenarios are the reasons why Equipment / Loads fry when higher than rated Voltage is applied to them.
Scott
