You need to know the current flow through the circuit as it is now, or the resistance of the circuit to be able to determine this.
If loading is very small (you're not drawing much current), you can set up a voltage divider so you get 1 volt across one resistor; this will waste electricity, so I don't suggest it if that is a serious concern. The resistors must be sized small relative to the current load to avoid causing problems with the circuit.
1 amp.
A 15 amp circuit breaker should trip at 15 amps regardless of the load voltages or impedances. If you have 277 volts and 7 ohms, the current would be 39.5 amps and a 15 amp circuit breaker should trip.
V = I*RIf you solved this for resistance, this means you would have:V/I = RYou set V = .9 volts, I = 1 amp, and solve to get .9 Ohms.
It's not that simple. The basic formula is Volts / Ohms = Amps. For 30 Volts you'd get 0.5 Amps, for 60 Volts you'd get 1 Amp, for 120 Volts you'd get 2 Amps.
R = voltage drop divided by current 13-6 = 7 volts, 7/1 is 7 ohms. If you have a device that has inrush (e.g., a motor or coil) you could have short-lived spikes well over triple the current, making it likely the voltage will drop during startup, which could damage the device. Personally, I would use a voltage divider to remove current from the equation: R2 = R1/ (v.in/v.out - 1) 0.86 x R2 = R1 (e.g., R2 is 1000 ohms and R1 is 860 ohms)
It's not that simple. The basic formula is Volts / Ohms = Amps. For 30 Volts you'd get 0.5 Amps, for 60 Volts you'd get 1 Amp, for 120 Volts you'd get 2 Amps.
amp*volts=watts
1 amp.
A 15 amp circuit breaker should trip at 15 amps regardless of the load voltages or impedances. If you have 277 volts and 7 ohms, the current would be 39.5 amps and a 15 amp circuit breaker should trip.
V = I*RIf you solved this for resistance, this means you would have:V/I = RYou set V = .9 volts, I = 1 amp, and solve to get .9 Ohms.
It's not that simple. The basic formula is Volts / Ohms = Amps. For 30 Volts you'd get 0.5 Amps, for 60 Volts you'd get 1 Amp, for 120 Volts you'd get 2 Amps.
Volts. Using the equation V=IR Change in voltage (measured in volts) = Current (in Amps) * Resistance (in ohms) So a volt equals amps times ohms.
The Orion CS 100.2 amp is rated for 100 watts RMS per channel at 2 ohms.
Ohm's Law states Volts = Amps x Resistance. You would need to apply 600 volts across 3 ohm load to have 200 Amps flow in circuit. Not sure what you are really asking and why you mentioned 2 gauge.
Ohm's law: Voltage is resistance times current. 80 ohms time 0.5 amperes = 40 volts.
The bulb you remove will go out :) Overall current will also be reduced proportional to the resistance of the bulb being removed. Lets say you have two 60 W incandescent bulbs in parallel and they each are drawing 1/2 Amp (60W = 120 Volts x 1/2 Amp). The resistance of each bulb is 240 Ohms (120 Volts / .5 Amps). The parallel resistance is 120 Ohms so 1 Amp is being drawn. When one of the two bulbs is removed the resistance changes from 120 Ohms to 240 Ohms, reducing the current from 1 Amp to 1/2 Amp.
Voltage drop is the decrease in electrical potential energy of electrons as they move through a circuit due to resistance. When electrons encounter resistance, they transfer some of their energy to overcome it, resulting in a decrease in voltage along the circuit. This drop in voltage is proportional to the resistance in the circuit and can affect the performance of electrical components.