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consider a light bulb with a filament made from tungsten (with a resistivity temperature coefficient of 4.5e-3/k). suppose the bulb is off, and at a temperature of 23 degrees c. then let the bulb be switched on. for the first instant, while the temperature has not yet had a chance to increase, the current in the bulb is 1.97 a. then the current decreases with time. a minute later, the current is only 0.64 a. what is the temperature t (in degrees c) of the filament at that time? the voltage driving the lightbulb filament stays the same as the filament heats.

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The temperature of the filament after a minute (t) is approximately 803°C.



To solve for the temperature of the filament (t) after a minute, we can follow these steps:

**1. Relate initial current (I0) to resistance (R0):**

- I0 = 1.97 A (given)

- V = constant voltage (given)

- R0 = V / I0 ≈ 5.08 Ω (calculated using Ohm's Law)

**2. Consider the change in resistance due to temperature increase:**

- ΔR = α * R0 * ΔT (where α is the resistivity temperature coefficient and ΔT is the temperature change)

- We know R0 from step 1 and α is given (4.5 * 10^(-3) K^(-1)), but ΔT is unknown.

**3. Apply conservation of energy to relate current change to temperature increase:**

- Power dissipated by the filament at initial and final states:

- P0 = I0^2 * R0

- P1 = I1^2 * R1 (where I1 is the current after 1 minute, 0.64 A)

- Since the voltage is constant, the power input remains the same (P0 = P1).

- Substitute R1 = R0 + ΔR and solve for ΔT:

- ΔT ≈ (P0 - P1) / (α * R0^2)

**4. Calculate the final temperature (t):**

- t = T0 + ΔT (where T0 is the initial temperature, 23 °C)

**Calculation:**

1. P0 = (1.97 A)^2 * 5.08 Ω ≈ 19.68 W

2. P1 = (0.64 A)^2 * 5.08 Ω ≈ 2.05 W

3. ΔT ≈ (19.68 W - 2.05 W) / (4.5 * 10^(-3) K^(-1) * 5.08 Ω^2) ≈ 780 K

4. t = 23 °C + 780 K ≈ 803 °C

**Therefore, the temperature of the filament after a minute (t) is approximately 803°C.**

User Mrpasqal
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