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Find the equivalent resistance of the circuit of the previous problem between the ends of an edge such as a and b in figure. |
Answer» Solution :Suppose a current `i` enters the circuit at the point a and the same current leaves the circuit at the point b. The current distribution is shown is figure.  The PATHS through .AD. and ah are equivalent and carry equal current `i_(1)`. The current through ab is then `i-2i_(1)`. The same distribution holds at the junction b. Currents in eb and cb are `i_(1)` each. The current `i_(1)` in ah is divided into a part `i_(2)` in he and `i_(1)-i_(2)` in hg. Similar is the division of current `i_(1)` in ad into dc and dg. The rest of the currents may be written easily using Kirchhoff.s junction law. The potential difference V between a and b may be written from the paths ab, aheb and ahgfcb as `V=(i-2i_(1))R, V=(i_(1)+i_(2)+i_(1))r` and `V=[i_(1)+(i_(1)-i_(2))+2(i_(1)-i_(2)+i(i_(1)-i_(2))+i_(1)]r` Which may be written as `V=(i-2i_(1))r, V=(2i_(1)+i_(2))r and V=(6i_(1)-4i_(2))r` Eliminating `i_(1) and i_(2)` from these equations, `(V)/(i)=(7)/(12)r` `therefore"The equivalent resistance "=7r//12` |
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