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six equal resistors, each of value R, are joined together as shown in the given figure 3.73. Calvulate the equvalant risistance accros AB. If a supply of emf E is connected accros AB, compute the current through the arm DE and AB.
question image
Asked by chrspranshu786 | 28 Jun, 2019, 10:40: PM
answered-by-expert Expert Answer
 
Let us assume potential at A is ε V and at B is 0 V .
If we measure the potential difference between A and F through the path A-C-D-F, we get ε V.
Since in this path A-C-D-F, we have two equal resistors, potential at D is ( ε/2 ) V.
 
Similarly if we measure the potential between A and H, through the path A-G-E-H, we get ε V.
Since in this path A-G-E-H, we have two equal resistors, potential at E is ( ε/2 ) V.
 
Hence D and E are at same potential, there will not be any current flow through the resistance
connected betwen D and E.
 
Hence this resistance between D and E can be removed. After removing the resistance,
remaining circuit is redrawn at right side of figure.
 
From the circuit drawn in right side, it can be easily worked out that equivalent resistance is R/2.
 
Current passing from A to B is  ε/(R/2)  or  2ε/R Ampere
Answered by Thiyagarajan K | 29 Jun, 2019, 02:25: PM

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