Practice portal › Current Electricity › Combination of Resistors
Asked in RS Academy GUJCET booklet · Symmetric and complex networks
Idea: reduce the ladder from the far end, where only simple series and parallel pairs appear, and work back towards A and B.
The top rail is all one node, A. At the far end, 3 Ω (F to A) in series with 3 Ω (E to F) gives 6 Ω from E to A.
That is in parallel with the 6 Ω rung from E to A: (6×6)/(6+6)=3 Ω.
Add 3 Ω (D to E) to get 6 Ω, again in parallel with a 6 Ω rung: 3 Ω from D to A.
The same step at C gives 3 Ω from C to A.
Add 3 Ω (B to C): 6 Ω from B to A through the ladder.
This is in parallel with the 3 Ω resistor joining A and B directly: R_AB=(6×3)/(6+3)=2 Ω.
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