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A battery-powered light bulb has a tungsten filament. When the switch connecting the bulb to the battery is first turned on and the temperature of the bulb is $20^{\circ} \mathrm{C},$ the current in the bulb is 0.860 A. After the bulb has been on for 30 s, the current is 0.220 A. What is then the temperature of the filament?
the temperature of the filament is $666.5^{\circ} \mathrm{C}$
Physics 102 Electricity and Magnetism
Chapter 19
Current, Resistance, and Direct-Current Circuit
Electric Charge and Electric Field
Gauss's Law
Electric Potential
Capacitance and Dielectrics
Current, Resistance, and Electromotive Force
Direct-Current Circuits
Electromagnetic Induction
Rutgers, The State University of New Jersey
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Current through the filament is 0.860 ampere at 20 degrees celsius after 30 seconds current through the filament is 0.220 ampere. Now we want to calculate a temperature at which current flowing through the filament is 0.20 ampere. So here we have current at 20 degree celsius is equal to 0.86 ampere. Here we have the reference temperature equal to 20 degrees celsius. Now, at latter temperature t we have an at that temperature equal to 0.220 ampere. Let v is the potential difference applied across the filament. Now the resistance of flament at 20 degree centigrade is equal to voltage applied across current flowing through the filament here. Voltage applied is v and gurn flowing through the filament is 0.60 ampere. So we have resistance equal to v upon 0.860 ampere. Now the resistance at t degrees celsius is equal to v upon current at 50 degree celsius that is equal to v. Upon current at d degree celsius is 0.220 ampere. Now we're using the relation that or a temperature t is equal to resistance at reference temperature into 1 plus temperature coefficient of resistance into temperature minus difference temperature. Now, why are putting different values here here? Residence at temperature t is v upon 0.220 ampere and resistance. At reference, temperature is v upon 0.60 ampere into 1 plus, and the value of alpha of dunstan flament is 0.0045 per degree centigrade into t. Minus 8 nat is 20 degree. Celcius from here, we can cancel b from both sides of the equation, so we have 0.860 ampere upon 0.220 ampere equal to 1 plus 0.0045 percent integratio t minus 20 degree celsius from here. We have 3.9090 equal to 1 plus 0.0045 per degree celsius into t minus 20 degree celsius. 93.9090. Minus 1 is equal to 0.0045 per degree celsius into t minus 20 degree celsius by subtracting the 2 terms. On the left hand, side of the equation we get 2.9090 is equal to 0.0045 per degree celsius into t minus 20 degree celsius. By dividing both sides by 0.0045 we get 2.9090 upon 0.0045 per degree. Celsius is equal to t minus 20 celcious from here we have t minus 20 degree. Celsius is equal to 646 degrees celsius or we have t equal to 646 degrees celsius plus 20 degrees celsius. So we have t equal to 6 s 6 degree. It'S the final answer:
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