a wire of resistance 20 ohm is folded to make a rectangle having side in the ratio 2 ratio 3 calculate the net resistance across the longer side of the rectangle
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On bending 20 ohms resistance into the square. At the ends of the opposite vertexes /diagonal, thise two points will devide the resistance into 2 equal resistances of 10 ohms each. These are the now in the parallel order. The total /final resistance will be 5 ohms. 1/R=1/R1 +1/R2
=1/10 +1/10
=2/10 ohms, R=5 ohms.
The electrical resistance of a wire would be expected to be greater for a longer wire, less for a wire of larger cross sectional area, and would be expected to depend upon the material out of which the wire is made.The resistance of a wire can be expressed as R=ρAL,
where,
ρ - Resistivity - the factor in the resistance which takes into account the nature of the material is the resistivity
L - Length of the conductor
A - Area of cross section of the conductor.
From this relation, we observe that the length is directly proportional to the resistance and the area of cross section is inversely proportional to the resistance.
That is, if L becomes 2 L, R becomes 2 R. R′=ρA2L. So, R = 2R'.
Hence, If the length of a wire is doubled, then its resistance becomes 2 times.
The electrical resistance of a wire would be expected to be greater for a longer wire, less for a wire of larger cross sectional area, and would be expected to depend upon the material out of which the wire is made.The resistance of a wire can be expressed as R=ρAL,
where,
ρ - Resistivity - the factor in the resistance which takes into account the nature of the material is the resistivity
L - Length of the conductor
A - Area of cross section of the conductor.
From this relation, we observe that the length is directly proportional to the resistance and the area of cross section is inversely proportional to the resistance.
That is, if L becomes 2 L, R becomes 2 R. R′=ρA2L. So, R = 2R'.
Hence, If the length of a wire is doubled, then its resistance becomes 2 times.