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what is resistivity value of filament in an electric bulb​

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An electric bulb has a power rating of 150 W. If the resistance of the filament in the resistor is 2 ohm what is the current flowing through it when connected to a 120 V source

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WILSON2014 eNotes educator | CERTIFIED EDUCATOR

The power rating tells us the amount of power that a resistor can safely dissipate as either heat or light without it breaking. In this case, the maximum power that the light bulb filament (resistor) can handle is 150W. The following are some useful formulas since we know that there is a 2 Ohm resistance and a 120V voltage source:

The power rating tells us the amount of power that a resistor can safely dissipate as either heat or light without it breaking. In this case, the maximum power that the light bulb filament (resistor) can handle is 150W. The following are some useful formulas since we know that there is a 2 Ohm resistance and a 120V voltage source:1. Voltage (V) = Current (I) x Resistance (R)

The power rating tells us the amount of power that a resistor can safely dissipate as either heat or light without it breaking. In this case, the maximum power that the light bulb filament (resistor) can handle is 150W. The following are some useful formulas since we know that there is a 2 Ohm resistance and a 120V voltage source:1. Voltage (V) = Current (I) x Resistance (R)2. Power (P) = Voltage (V) x Current (I) = I^2 x R

The power rating tells us the amount of power that a resistor can safely dissipate as either heat or light without it breaking. In this case, the maximum power that the light bulb filament (resistor) can handle is 150W. The following are some useful formulas since we know that there is a 2 Ohm resistance and a 120V voltage source:1. Voltage (V) = Current (I) x Resistance (R)2. Power (P) = Voltage (V) x Current (I) = I^2 x RFrom formula one, we can solve for the amount of current that passes through, where V=120V and R=2 Ohms. This gives I=60 Amps.

The power rating tells us the amount of power that a resistor can safely dissipate as either heat or light without it breaking. In this case, the maximum power that the light bulb filament (resistor) can handle is 150W. The following are some useful formulas since we know that there is a 2 Ohm resistance and a 120V voltage source:1. Voltage (V) = Current (I) x Resistance (R)2. Power (P) = Voltage (V) x Current (I) = I^2 x RFrom formula one, we can solve for the amount of current that passes through, where V=120V and R=2 Ohms. This gives I=60 Amps.This tells us that with this relatively small resistance value and the high voltage source, a relatively large current of 60 Amps are allowed to flow through. If we input these values to formula two (V=120V and I=60Amps), we find that the power output is 7200W!

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