Light bulbs are often assumed to obey Ohm\'s law. However, this is not really tr

Question

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.

Explanation / Answer

This is a discharge capacitor question, v_C = V_I e (-t/t) where: v_C = capacitor voltage at an instance in time in volts. V_I = voltage on capacitor at start of discharge cycle in volts. t (tau) = RC, time constant in seconds. t = time in seconds. R_filament = R = 3V/0.5A = 6O t (tau) = RC = 6O * 3F = 18s. It will take 5 time constants to discharge this capacitor completely or 5 * 18s = 90s. We assume time starts when light bulb is turned on or t = 0. Solve discharge formula for t. v_C = V_I e (-t/t) v_C/V_I = e (-t/t) Take natural log of both sides. -t/t = ln (v_C/V_I) t = - t ln (v_C/V_I) = - 18s * ln (2V/3V) = 7.29s So t_bright = 7 seconds.

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