A wire of resistance $ R $ is bent into a triangular pyramid as shown in the figure, with each segment having the same length. The resistance between points $ A $ and $ B $ is $ \frac{R}{n} $. The value of $ n $ is:
A wire of resistance $ R $ is bent into a triangular pyramid as shown in the figure, with each segment having the same length. The resistance between points $ A $ and $ B $ is $ \frac{R}{n} $. The value of $ n $ is:
Show Hint
- 16
- 14
- 10
- 12
The Correct Option is D
Approach Solution - 1
We are given a triangular pyramid formed by a wire with resistance \( R \), and each segment has the same length.
The wire is bent into a pyramid, and we are asked to find the resistance between points \( A \) and \( B \).
Since the resistance between \( A \) and \( B \) is given as \( \frac{R}{n} \), we need to analyze the resistances between the points of the pyramid.
The resistance between any two points in a complex circuit like this one can be found by combining the individual resistances of each segment in parallel and series.
We first note that the pyramid is symmetric, and each leg of the pyramid forms a resistance path. Using the symmetry of the pyramid and the principle of parallel and series resistances, we can derive the value of \( n \).
After solving the circuit using the properties of resistances in parallel and series, we find that the value of \( n \) is \( 12 \).
Thus, the resistance between points \( A \) and \( B \) is \( \frac{R}{12} \). %
Final Answer n = 12
Approach Solution -2
Given that: \[ r = \frac{R}{6} \] Since a balanced Wheatstone bridge is formed, the resistance between points \(A\) and \(B\) can be simplified as shown. The equivalent resistance between \(A\) and \(B\) is given by: \[ \frac{1}{R_{AB}} = \frac{1}{2r} + \frac{1}{2r} + \frac{1}{r} \] Simplify: \[ \frac{1}{R_{AB}} = \frac{2}{r} \] \[ R_{AB} = \frac{r}{2} \] Now substituting \(r = \frac{R}{6}\), \[ R_{AB} = \frac{R}{6 \times 2} = \frac{R}{12} \] \[ \boxed{R_{AB} = \frac{R}{12}} \]
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