Series and Parallel Arrangements of Resistances

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Solved Problems from IIT JEE

Problem from IIT JEE 2004

Six equal resistances are connected between points P, Q and R as shown in figure. Then, the net resistance will be maximum between,

Six equal resistances are connected between
  1. P and Q
  2. Q and R
  3. P and R
  4. any two points

Solution: The equivalent resistance between P and Q, between Q and R, and between P and R are given by \begin{alignat}{2} &R_\text{PQ}=R\parallel(R/2+R/3)=R\parallel 5R/6=\frac{R (5R/6)}{R+5R/6}=\frac{5}{11}R,\nonumber\\ &R_\text{QR}=(R/2)\parallel(R+R/3)=\frac{4}{11}R,\nonumber\\ &R_\text{PR}=(R/3)\parallel(R+R/2)=\frac{3}{11}R.\nonumber \end{alignat}

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References

  1. IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
  2. Collision Theory (MIT, pdf)