Question

What is the approximate resistance of a 100 W lightbulb if the AC voltage provided to it is given by v(t) = 200√2 cos(100πt)? R= __________Ω

Answer 1

The approximate resistance of a 100 W lightbulb with a given AC voltage of v(t) = 200√2 cos(100πt) is 400 ohms.

Step-by-step explanation:

The resistance R of a 100 W lightbulb with an AC voltage v(t) = 200√2 cos(100πt) can be calculated using the power formula, P = V2/R, where P is the power in watts, V is the RMS (Root Mean Square) voltage, and R is the resistance in ohms (Ω).

First, determine the RMS voltage. The given voltage is in the form of v(t) = V0cos(100πt), where V0 = 200√2 V is the maximum voltage. For a cosine function, the RMS voltage is VRMS = V0/√2, hence VRMS = 200 V.

Using the power formula, we get R = V2/P = (200 V)2/100 W = 400 Ω.

Therefore, the approximate resistance of the lightbulb is 400 Ω.