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A wire of resistance 18Ohm is bent to form an equilateral triangle. The resistance across any two vertices of the equilateral triangle is (in Ohm):
A. 12
B. 6
C. 4
D. 9

Answer
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Hint: Draw an equivalent circuit diagram of the given situation to determine the resistance across any two vertices of the equilateral triangle. Use the formula for the equivalent resistance for the resistors in series and parallel.

Formula used:
The equivalent resistance of the two resistors connected in series is
RS=R1+R2
Here, RS is the equivalent resistance, R1 is the resistance of the first resistor and R2 is the resistance of the second resistor.
The equivalent resistance of the two resistors connected in parallel is
1RP=1R1+1R2
Here, RP is the equivalent resistance, R1 is the resistance of the first resistor and R2 is the resistance of the second resistor.

Complete step by step answer:
The equilateral triangle is a triangle in which all of its three edges are of the same length.
The total resistance across any wire is directly proportional to the length of the wire.
The total resistance of the given wire is 18Ohm.
The given wire is bent to form an equilateral triangle. Hence, the total resistance 18Ohm of the wire is divided equally in three edges of the equilateral triangle.
Hence, the resistance through each edge of the equilateral triangle is 18Ohm3=6Ohm.
Now, calculate the resistance across any two vertices of the equilateral triangle.
The resistance across any two vertices of the equilateral triangle should be different as across any two vertices of the equilateral triangle two resistances are in series.
The diagram representing the resistances across 3 edges of the equilateral triangle and its equivalent circuit is as follows:
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From the above diagram, it is clear that when a wire with resistance is bent into a equilateral triangle, its resistance divides equally in three edges of triangle and the its equivalent circuit across two vertices of the triangle contains resistance of the two edges of the triangle in series and resistance of one edge in parallel to them.
Let the two resistors in series be labeled R1 and R2 the third resistor in parallel to them is labeledR3.
Calculate the equivalent resistance RS of the two resistances in series.
Substitute 6Ohm for R1 and 6Ohm for R2 in equation (1).
RS=(6Ohm)+(6Ohm)
RS=12Ohm
Hence, the equivalent resistance across the two parallel resistors is 12Ohm.
Now, calculate the equivalent RP resistance across the parallel resistance 12Ohm and
6Ohm.
Rewrite equation (2) for the equivalent resistance of the parallel resistances.
1RP=1RS+1R3
Substitute 12Ohm for RS and 6Ohm for R3 in the above equation.
1RP=112Ohm+16Ohm
1RP=6Ohm+12Ohm(12Ohm)(6Ohm)
RP=4Ohm
Therefore, the equivalent resistance across any two vertices of the equilateral triangle is 4Ohm.

So, the correct answer is “Option C”.

Note:
One can also draw the equivalent circuit diagram with resistance of one edge in the upper arm and of two edges (in series) in the lower arm and do the needful calculations to determine the equivalent resistance. Both methods will result with the same option.