No Problemmo #10

An electrical network of wires is formed by joining the grid points of two hexagons, as shown, where the resistance of each side and the connecting wires is R . R. An ideal battery of emf E E is joined at the bottom.

Considering that the wire connected with the cell is ideal, find the current passing through the battery.

Take E = 11 V E= 11\text{ V} and R = 6 Ω . R= 6\,\Omega.


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The answer is 3.

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2 solutions

João Areias
Jun 21, 2018

Consider these currents on the loops

By kirchoff's voltage law (closed loops) we have that:

[ R R 0 0 0 0 0 R 4 R R 0 0 0 R 0 R 4 R R 0 0 0 0 0 R 4 R R 0 0 0 0 0 R 4 R R 0 0 0 0 0 R 4 R R 0 R 0 0 0 R 4 R ] [ I 1 I 2 I 3 I 4 I 5 I 6 I 7 ] = [ V 0 0 0 0 0 0 ] \left[ \begin{array}{ccccccc} R&-R&0&0&0&0&0\\ -R&4*R&-R&0&0&0&-R\\ 0&-R&4*R&-R&0&0&0\\ 0&0&-R&4*R&-R&0&0\\ 0&0&0&-R&4*R&-R&0\\ 0&0&0&0&-R&4*R&-R\\ 0&-R&0&0&0&-R&4*R \end{array} \right]\cdot \left[ \begin{array}{c} I_1\\ I_2\\ I_3\\ I_4\\ I_5\\ I_6\\ I_7 \end{array} \right] = \left[ \begin{array}{c} V\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0 \end{array} \right] Solving this with your favorite method, mine is Cramer's rule, and we get that I 1 = 2.578125 I_1 = 2.578125 or I 1 3 I_1 \approx 3 rounding to the nearest integer.

Rounding the answer of 330 113 2.92035398230089 \frac{330}{113}\approx 2.92035398230089 to the nearest integer gives 3. The problem ought to have specified that rounding. The problem was solved by inverting an appropriate Kirchhoff's Laws derived matrix.

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