I need help on this 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 A when connected to a 3-V voltage source. For this problem, assume that the changing resistance causes the current to be 0.5 A for any voltage between 2 and 3 V.
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 F (an unusually large but not unrealistic value) and is initially charged to 3 V, how long will it take for the voltage across the flashlight bulb to drop to 2 V (where the bulb will be orange and dim)? Call this time tbright.
Express t_bright numerically in seconds to the nearest integer.
t_bright = 12 seconds
To solve this problem, we need to determine how the voltage across the flashlight bulb changes over time when connected to the given capacitor. We can use the concept of time constant, which represents the time it takes for the voltage across a capacitor to change by approximately 63.2% of its initial value.
The time constant (τ) is given by the equation: τ = R * C, where R is the resistance and C is the capacitance. In this problem, we are given the capacitance of the capacitor as 3 F.
The resistance of the flashlight bulb is not constant, but we are told that it draws a current of approximately 0.5 A for any voltage between 2 V and 3 V. To find an equivalent resistance that would give us the same current of 0.5 A at 3 V, we can use Ohm's law: R = V / I.
So, R = 3 V / 0.5 A = 6 Ω.
Now that we have the resistance (R) and capacitance (C), we can calculate the time constant (τ):
τ = R * C = 6 Ω * 3 F = 18 s.
The time constant represents the time it takes for the voltage across the flashlight bulb to decrease to approximately 37% of its initial value. In other words, after one time constant (τ), the voltage across the bulb drops from 3 V to 3 V * 0.368 = 1.104 V.
To find the time it takes for the voltage across the bulb to drop from 3 V to 2 V (i.e., from full brilliance to orange and dim), we can set up the following equation:
2 V = 3 V * e^(-t/τ),
where t is the time we want to find.
Rearranging the equation, we get:
e^(-t/τ) = 2/3.
Taking the natural logarithm (ln) of both sides:
-t/τ = ln(2/3).
Solving for t:
t = -τ * ln(2/3).
Plugging in the value for τ, we can calculate:
t = -18 s * ln(2/3) ≈ 6.24 s.
Rounding to the nearest integer, the time it takes for the voltage across the flashlight bulb to drop to 2 V is approximately 6 seconds.
To solve this problem, we can use the equation that relates voltage, current, and time for a capacitor discharge. The equation is:
V = V₀ * e^(-t/RC)
where:
- V is the voltage across the capacitor at time t
- V₀ is the initial voltage across the capacitor
- e is the base of the natural logarithm (approximately 2.71828)
- t is the time in seconds
- R is the resistance (changing in this case due to the filament heating up)
- C is the capacitance
In this case, we want to find the time it takes for the voltage across the flashlight bulb to drop to 2 V. The initial voltage is 3 V, the final voltage is 2 V, the capacitance is 3 F, and the resistance is not provided directly.
Since the problem states that the current through the bulb is 0.5 A for any voltage between 2 and 3 V, we can assume that the resistance is constant at any voltage within this range. Therefore, we can use Ohm's Law to derive the resistance:
R = V / I = 3 V / 0.5 A = 6 Ω
Now we can substitute the values into the equation and solve for t:
2 V = 3 V * e^(-t / (6 Ω * 3 F))
To solve for t, we need to isolate it. Starting by dividing both sides by 3 V:
2/3 = e^(-t / (6 Ω * 3 F))
Taking the natural logarithm (ln) of both sides to remove the exponential:
ln(2/3) = -t / (6 Ω * 3 F)
Now, we can solve for t by isolating it:
t = -ln(2/3) * (6 Ω * 3 F)
Using a calculator, we can find the value of t:
t ≈ -ln(2/3) * (6 Ω * 3 F) ≈ 3.1976 s
Rounding to the nearest integer, the time it takes for the voltage across the flashlight bulb to drop to 2 V is approximately 3 seconds.