Light bulbs are often assumed to obey Ohm's law. However, this is not really true because their resistance increases substantially as the filament heats up in its "working" state. A typical flashlight bulb at full brilliance draws a current of approximately 0.5 when connected to a 3- voltage source. For this problem, assume that the changing resistance causes the current to be 0.5 for any voltage between 2 and 3 . Suppose this flashlight bulb is attached to a capacitor as shown in the circuit from the problem introduction. If the capacitor has a capacitance of 3 (an unusually large but not unrealistic value) and is initially charged to 3 , how long will it take for the voltage across the flashlight bulb to drop to 2 (where the bulb will be orange and dim)? Call this time . Express numerically in seconds to the nearest integer.

   t = 7 10⁻⁶ s

Explanation:

For this exercise the bulb, the voltage source and the capacitor are connected in series. With the initial values let's look for resistance

        V = I R

         R = V / I

         R = 3 / 0.5

         R = 6 Ω

Let's look for the circuit time constant

         τ = R C

         C = 3 μF = 3 10⁻⁶ F

         τ = 6 3 10⁻⁶

        Tau = 18 10⁻⁶ s

The charge on this capacitor is

            Q = C V

            Q = 3 10⁻⁶  3

            Q = 9 10⁻⁶ C

The expression for capacitor discharge is

            I = -Q /RC  

           V = I R

           V = - Q /C  e^{-t/RC}

Let's replace

          V = - 9 10⁻⁶/3 10⁻⁶    e^{-t/RC}

          V = -3 e^{-t/RC}

          t /τ = -ln (V / 3)

          t = -τ  ln (V / 3)

Let's calculate

          t = -18 10⁻⁶ ln (2/3)

          t = 7.3 10⁻⁶ s

          t = 7 10⁻⁶ s

it doesn't. it was imagined up in the middle ages.