JEE Physics/ Lakshmikanta Satapathy/ Direct current/ Question on Resistance across one side of a cube made from twelve identical wires of given resistance solved using the symmetry of the network

1. Physics Helpline
L K Satapathy
Direct Current - 3
Resistance Cube 3
Resistance across one edge of a Cube

2. Physics Helpline
L K Satapathy
Twelve identical wires , each of
resistance 60 ohms , are
connected together to form a
cube ABCDEFGH as shown in
the figure. Find the resistance
between the points A and H of
the cube. Also find the resistance
between A and H when the wire
AH is removed.
A
D C
B
E F
GH
Direct Current - 3
Resistance across one edge of a Cube

3. Physics Helpline
L K Satapathy
Direct Current - 3
Resistance across one edge of a Cube
I2
I2-I3
I2
I3
I1
I
I3
I2
I2
2(I2-I3)
A
D C
B
E F
GH
I2-I3
V
Let IAB = IAD = I2
Let us connect a cell of emf V volts across AH
Let the current entering the network at A = I
Wires AB and AD are symmetric about AH
 They carry equal currents
Let the resistance of each wire = r [ given r = 60  ]
Let IAH = I1
Applying Kirchhoff’s junction rule at point A , we get
I = I1 + 2I2 - - - - - (1)

4. Physics Helpline
L K Satapathy
Direct Current - 3
Resistance across one edge of a Cube
Applying junction rule at D and B , we get IDC = IBC = (I2 - I3)  ICF = 2(I2 - I3)
Applying junction rule at E and G , we get IFE = IFG = (I2 - I3)
Let IDE = IBG = I3
 IGH = IEH = I2
Again wires GH and EH are symmetric about AH.
 They carry equal currents.
I2
I2-I3
I2
I3
I1
I
I3
I2
I2
2(I2-I3)
A
D C
B
E F
GH
I2-I3
V
Current I = I1 + 2I2 must leave the network at H
Again wires DE and BG are symmetric about AH.
 They carry equal currents.
Now we can define the currents in the other wires of the network.

5. Physics Helpline
L K Satapathy
Direct Current - 3
Resistance across one edge of a Cube
I2
I2-I3
I2
I3
I1
I3
I2
I2
2(I2-I3)
A
D C
B
E F
GH
I2-I3
Let the required resistance between A and H = x
And for the wire AH , V = I1 r
 (I1 + 2I2)x = I1 r - - - (2)
 I1 r = I2 r + I3 r + I2 r = (2I2 + I3)r
 I1 = 2I2 + I3 - - - - - (3)
 I3r = (I2-I3)r + 2(I2-I3)r + (I2-I3)r
 I3 = 4(I2-I3)  5I3 = 4I2  I3 = - - - - (4)
5
4I2
Applying Ohm’s law for the network between A and H
V = Ix = (I1 + 2I2)x [ using eqn. (1) ]
For loop ABGH , VAH = VAB + VBG + VGH
For loop BCFG , VBG = VBC + VCF + VFG

6. Physics Helpline
L K Satapathy
Direct Current - 3
Resistance across one edge of a Cube
1 2 3 2 2 2
2 1
4 14
(3) &(4) 2 2
5 5
5
14
I I I I I I
I I
     
 
1 2 1(2) ( 2 )I I x I r  
1 1 1 1 1
5 12
7 7
I I x I r I x I r
 
     
 
7 7
60 35
12 12
[ ]A sr nx     

7. Physics Helpline
L K Satapathy
Direct Current - 3
Resistance across one edge of a Cube
Resistance of cube about AH = x
Resistance of removed wire = r
Let resistance of remaining cube about AH = y
We need to find the value of y.
Resistance of y and r in parallel = x
1 1 1
y r x
  
1 1 1 12 1 5
7 7y x r r r r
     
7 7 60
84
5 5
[ ]
r
A sy n

    
A
D C
B
E F
GH
For the 2nd part of the question , wire AH is removed as shown.

8. Physics Helpline
L K Satapathy
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