91影视

Equivalent/Total resistance of two resistors ${\mathit{R}}_{\mathbf{1}}$and${\mathit{R}}_{\mathbf{2}}$ in series is given as:

${\mathit{R}}_{\mathbf{e}\mathbf{q}}\mathbf{=}{\mathit{R}}_{\mathbf{1}}\mathbf{+}{\mathit{R}}_{\mathbf{2}}$

Equivalent/Total resistance of two resistors ${\mathit{R}}_{\mathbf{1}}$and ${\mathit{R}}_2$in parallel is given as:

$\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{e}\mathbf{q}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{R}}_{\mathbf{2}}}$

Assume a resistorlies along each edge of a cube ( 12 resistors in all) with connections at the corners as shown in the following figure. Let the current enters from the one corner and exit from the opposite corner, as shown.

Since all the resistor have the sameresistance, then when the current enter it will be equally distributed over the threeresistors (in the first junction so each will have$II3$. In the next branch the current will beequally distributed again but this time over two resistors instead of three so each will get the half of$II3$, which is$II6$. Finally, the current will be added to the next level of resistorsuntil it exits the network. The current is shown in the figure.

The voltage drop from tothen is,

$$\begin{gathered} V=\left(\frac{1}{3}\right)R+\left(\frac{1}{6}\right)R=\left(\frac{5}{6}\right)IR \\ \mathrm{But},V=I{R}_{eq} \end{gathered}$$

Thus,

$R_{eq}=\frac{5}{6}R$