The corroded contacts in a lightbulb socket have 8.20 Omega resistance.

How much actual power is dissipated by a 105 W (120 V) lightbulb screwed into this socket?

The resistance of the bulb at operating temperature is R = V^2/P = 137.1 ohms. If it is in series with 8.2 ohm contact resistance, the voltage applied to the bulb filament is reduced to 120*(137.1/145.3) = 113.2 V, and the power dissipated INE THE BULB is reduced to V^2/R = (113.2)^2/R = 93.5 W. There will be additional power dissipated at the corroded contacts, equal to 6.8^2/8.2 = 5.6 W

To find the amount of actual power dissipated by the lightbulb in the socket, you can use Ohm's Law and the formula for power.

First, let's calculate the current flowing through the socket using Ohm's Law:

Ohm's Law: V = I * R

Where:
V is the voltage (120 V)
I is the current
R is the resistance (8.20 Ω)

Rearranging the formula to solve for I, we have:

I = V / R

Substituting the given values, we get:

I = 120 V / 8.20 Ω ≈ 14.63 A

Now that we have the current flowing through the socket, we can calculate the power dissipated by the lightbulb using the formula:

Power (P) = Current (I) * Voltage (V)

Substituting the given values, we have:

P = 14.63 A * 120 V
P ≈ 1755.6 W

Therefore, approximately 1755.6 W of actual power is dissipated by the 105 W (120 V) lightbulb screwed into this socket.