Electricty runs in a loop from neg to pos (in DC). From your question, it appears you're talking about AC. If it is single-phase, it still runs in a loop, so it *SEEMS* from your description that it would not be additive.
Best to draw a diagram of what you're measuring when you say "measured amps on both legs" and resubmit the question before I give you a 100% answer.
Watts is the product of amps times volts. Reduce either or both of these values will reduce the overall wattage output in a resistive circuit.
ANSWER: The brightness of both bulbs will decrease. If the bulbs are identical the current will decrease to 0.2 Amps. This is a simple series resistive circuit, the more bulbs you add in series both the amperage and bulb brightness will continue to go down.
There is no rating for #14 wire in the electrical code book. This is because #12 aluminium wire is rated at 20 amps and that is the minimum of home wiring circuit wiring in aluminium. It is not, if at all, used anymore. #10 at 75 or 90 degree C is rated both at 30 amps. #8 at 75 or 90 degree C is rated both at 45 amps.
form you question i found out that your circuit bearker panal has burn too ,,in case not to happne again you please have got to check the AMPS at both ,,the bulb and the hell circuit breaker ,,the buy the high of AMPS for ur circit breaker .
A 120 v circuit would supply 120 v to both resistors if they are in parallel, which is 120/100 amps into a 100 ohm load, and 120/80 amps into am 80 ohms load, which totals up to 2.7 amps, so the total power is 120x2.7 watts or 324 watts.
In both cases, the power dissipated is measured by multiplying the voltage across the circuit by the current through the circuit.
In both cases, the power dissipated is measured by multiplying the voltage across the circuit by the current through the circuit.
Yes. Power in both cases is 1.035 KW. Your meter shouldn't know the difference.
Watts is the product of amps times volts. Reduce either or both of these values will reduce the overall wattage output in a resistive circuit.
An ammeter is a amp meter put into a circuit in series. There is virtually no voltage drop or resistance in an ammeter so two in series would be redundant. If you have one in a circuit it will tell you the amps that circuit is generating, two would both give virtually the same result.
ANSWER: The brightness of both bulbs will decrease. If the bulbs are identical the current will decrease to 0.2 Amps. This is a simple series resistive circuit, the more bulbs you add in series both the amperage and bulb brightness will continue to go down.
There is no rating for #14 wire in the electrical code book. This is because #12 aluminium wire is rated at 20 amps and that is the minimum of home wiring circuit wiring in aluminium. It is not, if at all, used anymore. #10 at 75 or 90 degree C is rated both at 30 amps. #8 at 75 or 90 degree C is rated both at 45 amps.
You have to remember ohms law Voltage = amp * resistance. Using some basic algebra you can rewrite the equation as amps = voltage / resistance. Since a short circuit has relatively 0 ohms of resistance, this increases both the amps and resistance which uses more battery capacity,power, and creates more heat.
form you question i found out that your circuit bearker panal has burn too ,,in case not to happne again you please have got to check the AMPS at both ,,the bulb and the hell circuit breaker ,,the buy the high of AMPS for ur circit breaker .
KV is the unit used for VOLTAGE measurement. AC power is complex quantity that is it has both magnitude and direction and hence has two parts real part and imaginary part. complex power is measured in KVA (kilo volts amps) real part (active component ) is measured in KW (kilo watts) imaginary part (reactive component) is measured in KVAR (kilo volts amps reactive)
1300 watts on a 120 volt circuit is 10.8 amps. Since most circuits are built with 15 or 20 amp breakers, no. Rdrsh is correct. If both outlets you plan on using are on the same circuit and you have nothing else on, you might be able to run both at the same time for a short period of time if it is a 20 amp circuit. If it is a 15 amp circuit, no way will it work. However if you have a couple of 100 watt lights on then you will have a total of over 20 amps draw and it will trip the breaker. You need to have these of separate circuits.
This is most easily done using an oscilloscope with a current probe. Also phase angle must be measured relative to something else, usually the AC voltage at one end or the other of the branch of the circuit where the current is being measured.