Question:
Six equal resistances are connected between points P, Q and R as shown in figure. Then net resistance will be 
Six equal resistances are connected between points P, Q and R as shown in figure. Then net resistance will be
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Reduce the figure to three branches: P-Q is one r; Q-R is two r in parallel; P-R is three r in parallel. Compute the three pair resistances.
Updated On: Nov 4, 2026
- P and R
- P and Q
- Q and R
- Any two points
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The Correct Option is B
Solution and Explanation
From the figure: branch PQ = single \(r\); branch QR = two resistors in parallel \(= r/2\); branch PR = three resistors in parallel \(= r/3\).
Between P and Q: \(r\) in parallel with \(\left(\dfrac{r}{3} + \dfrac{r}{2}\right) = \dfrac{5r}{6}\): \(R_{PQ} = \dfrac{r\cdot 5r/6}{r + 5r/6} = \dfrac{5r}{11}\).
Between P and R: \(\dfrac{r}{3}\) in parallel with \(\left(r + \dfrac{r}{2}\right) = \dfrac{3r}{2}\): \(R_{PR} = \dfrac{3r}{11}\).
Between Q and R: \(\dfrac{r}{2}\) in parallel with \(\left(r + \dfrac{r}{3}\right) = \dfrac{4r}{3}\): \(R_{QR} = \dfrac{4r}{11}\).
The maximum is \(\dfrac{5r}{11}\) between P and Q.
Answer: (2) P and Q
Between P and Q: \(r\) in parallel with \(\left(\dfrac{r}{3} + \dfrac{r}{2}\right) = \dfrac{5r}{6}\): \(R_{PQ} = \dfrac{r\cdot 5r/6}{r + 5r/6} = \dfrac{5r}{11}\).
Between P and R: \(\dfrac{r}{3}\) in parallel with \(\left(r + \dfrac{r}{2}\right) = \dfrac{3r}{2}\): \(R_{PR} = \dfrac{3r}{11}\).
Between Q and R: \(\dfrac{r}{2}\) in parallel with \(\left(r + \dfrac{r}{3}\right) = \dfrac{4r}{3}\): \(R_{QR} = \dfrac{4r}{11}\).
The maximum is \(\dfrac{5r}{11}\) between P and Q.
Answer: (2) P and Q
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