JEE Physics/Lakshmikanta Satapathy/Problem on Resistance across the main diagonal of a cube made from twelve identical wires solved using the symmetry of the network

1. Physics Helpline
L K Satapathy
Direct Current - 1
Resistance Cube 1
Resistance across Main Diagonal of a Cube

2. Physics Helpline
L K Satapathy
Direct Current - 1
Resistance across Main Diagonal of a Cube
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 F of the
cube.
A
D C
B
E F
GH

3. Physics Helpline
L K Satapathy
Direct Current - 1
Resistance across Main Diagonal of a Cube
Let us connect a cell of emf V volts across AF
Let the Resistance of each wire = r
. . . (2)
3
AB AD AH
I
I I I   
. . . (3)
3
AB AD AH
Ir
V V V   
A
D C
B
E
F
GH
I
V
. . . . (1)V I x 
Branches AB , AD and AH are symmetric with respect
to AF. Hence current I is equally shared by them.
Also the resistances of AB , AD and AH are each equal to r
Let current I enter the network at A.
and Resistance between A and F = x
Applying Ohm’s Law between points A and F

4. Physics Helpline
L K Satapathy
Direct Current - 1
Resistance across Main Diagonal of a Cube
1B D HLet V V V V  
 Potentials of points B , D and H are equal.
. . . (4)
3
CF EF GF
I
I I I   
. . . (5)
3
CF EF GF
Ir
V V V   
Also branches CF , EF and GF are symmetric with
respect to AF.  Current I is equally shared by them.
Resistances of CF , EF and GF are each equal to r .
A
D C
B
E
F
GH
V
I

5. Physics Helpline
L K Satapathy
Direct Current - 1
Resistance across Main Diagonal of a Cube
1 2( )
BC BG DC DE HE HGV V V V V V
V V
     
 
BC BG DC DE HE HGI I I I I I     
2C E GLet V V V V  
 Potentials of points C , E and G are equal.
PD across the wires BC , BG , DC , DE , HE and HG
are equal. Their resistances are also equal. Hence the
same current flows through each of these.
A
D C
B
E
F
GH
V
I

6. Physics Helpline
L K Satapathy
Direct Current - 1
Resistance across Main Diagonal of a Cube
2
AB
BC BG
I
I I  
Current in AB is equally shared by wires BC and BG
6
BC BG DC DE HE HG
I
I I I I I I      
. . . (6)
6
BC BG DC DE HE HG
Ir
V V V V V V      
A
D C
B
E
F
GH
V
I
(2)
3
AB
I
I 
6
BC BG
I
I I  

7. Physics Helpline
L K Satapathy
Direct Current - 1
Resistance across Main Diagonal of a Cube
AB BG GFV V V V  
5
.
3 6 3 6
Ir Ir Ir Ir
I x    
5 5 60
50
6 6
[ ]
r
A sx n

    
For the path ABGF , we have
We put the values from equations (1) , (3) , (6) and (5)
A
D C
B
E
F
GH
V
I

8. Physics Helpline
L K Satapathy
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