A uniform wire of resistance 40ohm is bent into the form of a square. What is the effective resistance between any two diagonally opposite cornes of the square?
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Given :
Total Resistance of wire =40 ohms
As it is bent in the form of square:
AB=BC=CD=AD
R1=R2=R3=R4=40/4=10 ohms
Resistance Across each Diagonal AC :R
1/R=1/R¹+1/R²
R¹=R1+R2=10+10=20 ohms
R²=R3+R4=10+10=20 ohms
1/R=1/20+1/20=2/20
=1/10R
=10 ohms
Total Resistance of wire =40 ohms
As it is bent in the form of square:
AB=BC=CD=AD
R1=R2=R3=R4=40/4=10 ohms
Resistance Across each Diagonal AC :R
1/R=1/R¹+1/R²
R¹=R1+R2=10+10=20 ohms
R²=R3+R4=10+10=20 ohms
1/R=1/20+1/20=2/20
=1/10R
=10 ohms
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Your solution is as follows:
10 ohm.
When wire is bent in square shape the resistance at each side of square is same, so it is one fourth of the total resistance.
So resistance of each side is 10 ohm. We estimate the two contradictory diagonal so the two side of square is in succession and that series grouping is parallel with another series grouping
Series grouping has 10+10=20 ohm resistance and that 20 ohm resistance is parallel with another 20 ohm resistance. So, 20 !! 20 =10 ohm.
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