Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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Wheatstone Bridge Method
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The number of microstates (number of different ways to put the energy into the oscillators) for a specific macrostate (the total energy in the box) is called the multiplicity of that macrostate, and it is designated by Q. Thus, we could write 2(0) = 1 Q(1) = 3 2(2) = 6 2(3) = 10. The general formula for this is 2(N, q) (q+N-1)! q!(N-1)! where there are N oscillators with a total energy of q %3D units. Check your answers above using this formula. (There is no need to put any answers here.) We will assume for this that the energy can move among the three oscillators. Also, we say that the various microstates of the system (each ordered triplet) is equally probable of occurring for a given total energy value of the macrostate. (Just like dice rolls from the discussion class.) Let's look at how energy might be placed in a system with two equal pieces (two equal pieces of a chunk of metal). Here is our model: Let's have two identical boxes, labeled A and B, and each box has three oscillators. Thus NA = 3 and Ns = 3. We will allow these box be able to interact and exchange energy. We now place six units of energy into the system. Thus qtotal = qa + qe = 6. We could put all of it in the left and none in the right, or some in the left (A) and the rest in the right (B). Using the formula for the multiplicity of our system, N(N,q), complete the following table: qa(energy in A side) qolenergy in B side) 2 total = a 1 28 28 1 3 21 63 15 3 10 4 6 5 3
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Transcribed Image Text:The number of microstates (number of different ways to put the energy into the oscillators) for a specific macrostate (the total energy in the box) is called the multiplicity of that macrostate, and it is designated by Q. Thus, we could write 2(0) = 1 Q(1) = 3 2(2) = 6 2(3) = 10. The general formula for this is 2(N, q) (q+N-1)! q!(N-1)! where there are N oscillators with a total energy of q %3D units. Check your answers above using this formula. (There is no need to put any answers here.) We will assume for this that the energy can move among the three oscillators. Also, we say that the various microstates of the system (each ordered triplet) is equally probable of occurring for a given total energy value of the macrostate. (Just like dice rolls from the discussion class.) Let's look at how energy might be placed in a system with two equal pieces (two equal pieces of a chunk of metal). Here is our model: Let's have two identical boxes, labeled A and B, and each box has three oscillators. Thus NA = 3 and Ns = 3. We will allow these box be able to interact and exchange energy. We now place six units of energy into the system. Thus qtotal = qa + qe = 6. We could put all of it in the left and none in the right, or some in the left (A) and the rest in the right (B). Using the formula for the multiplicity of our system, N(N,q), complete the following table: qa(energy in A side) qolenergy in B side) 2 total = a 1 28 28 1 3 21 63 15 3 10 4 6 5 3
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