# 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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### 2 Answers

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

From 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!

The light bulb would not be able to handle this amount of power and would go out immediately when the circuit is closed. To prevent this from happening, we could either raise the value of the resistor or lower the voltage source so that the power never reaches the power rating. This example shows why some electrical appliances or devices may not necessarily work, even if we provide a source of electricity to them. In fact, it may even be very dangerous if we provide an inappropriate amount of power to a device as the example demonstrates.

**Sources:**

The power consumption of a resistor connected across a voltage V and with a current I flowing through it is P = I*V. When a voltage V is applied across a resistance R, the current that flows through it is I = V/R. Substituting V = I*R in the equation P = V*I gives P = I^2*R

It is given that the power rating of the electric bulb is 150 W, the resistance of the filament is 2 ohm and it is connected across a voltage source of 120 V. The power consumed is equal to 150 = 120*I where I is the current flowing through the resistor. This gives current as 1.25 A. But using P = I^2*R gives 150 = I^2*2 or I = `sqrt 75` A.

The different values of current obtained using the two equations indicates that the information provided is incorrect; the resistance of the filament of a bulb with the given power rating is not 2 ohm.