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Current Electricity question

2023 · 1 Feb · Shift 2 · Q50
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Current Electricity question

2023 · 1 Feb · Shift 2 · Q50

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance R would be :
  1. A
    (n−1)Rn2\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}^{2}}n2(n−1)R​
  2. B
    n2Rn−1\frac{n^{2} R}{n-1}n−1n2R​
  3. C
    (n−1)R(2n−1)\frac{(n-1) R}{(2 n-1)}(2n−1)(n−1)R​
  4. D
    (n−1)Rn\frac{(n-1) R}{n}n(n−1)R​
View written solutionFree

Correct answer: A

  1. Interpret the given resistance

A regular nnn-sided polygon is made of uniform wire of total resistance RRR.

Since all sides are equal, resistance of each side is r=Rn.r=\frac{R}{n}.r=nR​.

We need the equivalent resistance between two adjacent corners.

  1. Identify the two paths between adjacent corners

Take two adjacent vertices AAA and BBB.

Between AAA and BBB, current can go by two paths:

  • Direct side ABABAB: resistance r=Rnr=\frac{R}{n}r=nR​

  • Remaining (n−1)(n-1)(n−1) sides around the polygon: resistance (n−1)r=(n−1)Rn(n-1)r=(n-1)\frac{R}{n}(n−1)r=(n−1)nR​

These two paths connect the same endpoints AAA and BBB, so they are in parallel.

  1. Compute parallel combination

Thus, Req=r⋅(n−1)rr+(n−1)rR_{eq}=\frac{r\cdot (n-1)r}{r+(n-1)r}Req​=r+(n−1)rr⋅(n−1)r​

Simplify denominator: r+(n−1)r=nrr+(n-1)r=nrr+(n−1)r=nr

So, Req=(n−1)r2nr=n−1nrR_{eq}=\frac{(n-1)r^2}{nr}=\frac{n-1}{n}rReq​=nr(n−1)r2​=nn−1​r

Now substitute r=Rnr=\frac{R}{n}r=nR​: Req=n−1n⋅RnR_{eq}=\frac{n-1}{n}\cdot \frac{R}{n}Req​=nn−1​⋅nR​

Hence, Req=(n−1)Rn2.R_{eq}=\frac{(n-1)R}{n^2}.Req​=n2(n−1)R​.

  1. Match with options

The obtained result is (n−1)Rn2\boxed{\frac{(n-1)R}{n^2}}n2(n−1)R​​

This matches Option A.

  1. Comparison with stored answer

Stored correct answer: A

Derived answer: A

So they agree.

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