This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution 5: RC, Kirchhoff, EMF. TA: Yi Mao, email@example.com. 1. Exploding wire revisited: Capacitor has C c = 0.1 milliFarads; potential difference in capacitor is V 3500 Volts. Resistivity of nichrome wire is r = 1 . 5 10- 6 Ohm-meters; length of wire is L 20 centimeters; radius of wire is r 0.1 millimeters. Nichrome has a density m 8 . 4 10 3 kg/m 3 , and a specific heat C H 420 Joules/(kg-Kelvin). Before doing the problem, you should calculate the resistance of the nichrome wire R, its mass M, and its heat capacity (specific heat times mass). R = r L/ ( r 2 ) 10 Ohms ; M = m Lr 2 5 10- 5 kg ; c H = C H M . 02 Joules/Kelvin. (a) How much energy was stored in the capacitor before I discharged it through the nichrome wire? E = 1 / 2 C c V 2 600 Joules. (b) From the time evolution of the energy stored in an RC circuit, compute how much power was initially being dissipated in the wire. P | t =0 = I 2 R = V 2 /R 10 6 watt (c) Using the fact that the rate of a change of systems energy (dU/dt) is related to the rate of change of its temperature (dT/dt) by dU dt = ( C H M ) dT dt calculate the rate of change of the wires temperature from the initial power. dT dt | t =0 = P | t =0 / ( C H M ) = P | t =0 /c H 5 10 7 Kelvin/sec. (d) Nickel and chromium each boil near 3000 Kelvin. Given this, approximately how long did it take to evaporate the wire? (Note that our room temperature is about 300 Kelvin.) t T dT dt | t =0 5 10- 5 sec. This approximatin works because during this short time the voltage loss in the capacitor is not large, only about 5% of the original voltage. I I A B a a A B I I I 1 2 U A B R R (a) (c) (b) Figure 1: (a) Open circuit (b) Short-circuit at terminal A and B. (c) The Th evenin equivalent circuit. 2. Purcell 4.21: Th evenin equivalence of a circle of resistors and batteries. In the open circuit case, I a = 3 5 R = 0 . 009 ampere. U = U B- U A = 2 I a R- = 0 . 3 volts. (1) In the short-circuit case, we have U = U B- U A = 0, that is, 2 - 3 I 1 R = 0;- + 2 I 2 R = 0; I = I 1- I 2 . (2) The solution is I = 0 . 0025 ampere ....
View Full Document
- Spring '08