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Why there isn't current in R3? Does it depends on the resistance of R3?

  • 2
    \$\begingroup\$ It is a good exercise to assume that ONE of the four resistors R has another value (for example 1.1*R) and to calculate the current through R3. Then , you can see what happens when all are equal. \$\endgroup\$ – LvW May 17 '14 at 8:22
  • \$\begingroup\$ And better yet, try simulating it. \$\endgroup\$ – Dzarda May 17 '14 at 11:59
  • \$\begingroup\$ Is this homework or an assignment? That's allowed but if so you should say. \$\endgroup\$ – Russell McMahon May 17 '14 at 14:19

Its because there is no potential difference across the resistor \$R_3\$. It has nothing to do with the value of \$R_3\$.

  • \$\begingroup\$ As this seems likely to be a 'homework question' it may have been advisable to retain a degree of mystery :-). Asking them if it is may be usefuil. \$\endgroup\$ – Russell McMahon May 17 '14 at 14:18
  • \$\begingroup\$ @RussellMcMahon sorry about that. :-) just started answering. ll remember your advice while answering questions in future. \$\endgroup\$ – nidhin May 17 '14 at 15:21

The 5-resistor system is left-right symmetrical: if I exchange left and right, the current in \$R_3\$ changes its direction, but the circuit is the same (so the current remains the same, since there is only one possible value for this variable).

So it can be deduced that \$I = -I\$.


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