View source for Chapter 26 Problem 40

==Problem==
[[File:Chapter26problem40q.jpg|thumb|right|Resistance is futile!]]
Twelve resistors, each of resistance R, are connected as the edges of a cube. Determine the equivalent resistance 

(a) between points a and b, the ends of a side; 

(b) between points a and c, the ends of a face diagonal; 

(c) between points a and d, the ends of the volume diagonal.

[Hint: Apply an emf and determine currents; use symmetry at junctions.]

==solution==
[[File:Cube_symmetry.jpg|500px|center|Symmetries of a cube]]

===(a)===
[[File:Chapter26problem40s.jpg|300px|center|The symmetry reduced cube]]

Insert a probe battery between points (a) and (b). Let's label the current drawn from this probe battery as I

This setup has a symmetry plane between abed, so the current split at junction a will be <math>I=I_1+I_1+I_2</math>

Another symmetry is that the afhe plane is the mirror cdgb (since one has the current I entering, and the other has the current leaving) thus we have the same currents in erverse directions in those planes.

The 6 equations in the symmetry reduced cube is:
#<math>I=2I_1+I_2</math> 
#<math>I_3+I_4=I_1</math>
#<math>I_5=2I_4</math>
#<math>0=-2 I_1 R -I_3 R + I_2 R</math>
#<math>0=-2 I_4 R -I_5 R + I_3 R</math>
#<math>0=\mathcal{E}-I_2 R </math>

6 equations and 6 unknowns, so we start solving:

#<math>I=2(I_3+I_4) +\frac{\mathcal{E}}{R}</math> (2 and 6 in 1)
#<math>0=-2(I_3+I_4)R-I_3 R+\frac{\mathcal{E}}{R}R=-2 I_4 R-3 I_3 R+\mathcal{E}</math> (2 and 6 in 4)
#<math>0=-2 I_4 R-2 I_4 R+I_3 R=- 4 I_4 R + I_3 R</math> (3 in 5)


<math>I_3=4 I_4</math> ( solution of 3)

<math>0=-2 I_4 R - 3 (4 I_4) R+\mathcal{E} \rightarrow I_4=\frac{\mathcal{E}}{14R}</math> (previous into 2)

<math>I=2 (4 I_4)+ 2(I_4)+\frac{\mathcal{E}}{R}=10 I_4 + \frac{\mathcal{E}}{R}=\frac{10\mathcal{E}}{14R}+\frac{\mathcal{E}}{R}=\frac{12\mathcal{E}}{7R}</math> (previous into 1)

<math>R_{\textrm{eq}}=\frac{\mathcal{E}}{I}=\frac{7}{12}R</math>

===(b)===
[[File:Chapter26problem40s2.jpg|300px|center|The symmetry reduced cube]]

Insert a probe battery between points (a) and (c). Let's label the current drawn from this probe battery as I

This ehgb is a symmetry (inversion) plane afcd is a symmmetry plane.

#<math>I=2I_1 + I_2</math> (Junction rule at a)
#<math>2I_3=I_2</math> (Junction rule at d)
#<math>0=-2 I_2 R + \mathcal{E}</math> (loop rule around abca)
#<math>0=-2I_2 R -2 I_3 R + 2 I_1 R</math> (loop rule around afgdcha)

Then

<math>I_2 = \frac{\mathcal{E}}{2R}</math>

<math>I_3 = \frac{I_2}{2} = \frac{\mathcal{E}}{4R}</math>

<math> I_1 = I_2 + I_3 = \frac{\mathcal{E}}{2R} + \frac{\mathcal{E}}{4R}=\frac{3\mathcal{E}}{4R}</math>

<math>I=2I_1 + I_2=2\left(\frac{3\mathcal{E}}{4R}\right)+\frac{\mathcal{E}}{2R}=\frac {2\mathcal{E}}{R}</math>

<math>R_{\textrm{eq}}=\frac{\mathcal{E}}{I}=\frac {1}{2}R</math>

===(c)===
[[File:Chapter26problem40s3.jpg|300px|center|The symmetry reduced cube]]

Insert a probe battery between points (a) and (d). Let's label the current drawn from this probe battery as I

#<math>I=3I_1</math> (junction rule at a)
#<math>I_1=2I_2</math> (junction rule at b)
#<math>0=-2 I_1 R-I_2 R + \mathcal{E}</math> (loop rule abgda)

<math>0=-2 I_1 R-\frac {1}{2}I_1 R + \mathcal{E} \rightarrow I_1=\frac{2\mathcal{E}}{5R}</math>

<math>I=3I_1=\frac{6\mathcal{E}}{5R}</math>

<math>R_{\textrm{eq}}=\frac{\mathcal{E}}{I}=\frac {5}{6}R</math>